Coordinate Geometry-i Circles (Lecture-03)
Example 1
Find the equation of the image of the circle $x^{2}+y^{2}+16 x-24 y+183=0$ by the line mirror $4 x+7 y+13=0$
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Solution:
The given circle and line are
$\mathrm{x}^{2}+\mathrm{y}^{2}+16 \mathrm{x}-24 \mathrm{y}+183=0$______________(1) and $4 x+7 y+13=0$ _________________(2)
Centre and radius if the circle are
$(-8,12)$ and $\sqrt{64+144-183}=\sqrt{25}=5$ respectively.
Equation of line $\mathrm{C} _{1} \mathrm{C} _{2}$ is $7 \mathrm{x}-4 \mathrm{y}+\mathrm{k}=0$ it passes through $(-8,12)$
$\therefore-56-48+\mathrm{k}=0$
$\mathrm{k}=104$
Equation of line $\mathrm{C} _{1} \mathrm{C} _{2}$ is $7 \mathrm{x}-4 \mathrm{y}+104=0$
To get the coordinates of M. Solve the equation
(2) & (3)
(2) $\times 4+(3) \times 7$
$16 \mathrm{x}+28 \mathrm{y}+52=0$
$\begin{gathered} 49 x-28 y+728=0 \\ \hline 65 x+780=0 \end{gathered}$
$\mathrm{x}=-12$
put the value of $x$ in (2) we get
$-48+7 y+13=0$
$7 \mathrm{y}=35$
$\mathrm{y}=5$
$\therefore$ coordinate of $\mathrm{M}$ is $(-12,5)$
$M$ is the midpoint of $\mathrm{C} _{1}$ and $\mathrm{C} _{2}$
$\therefore-12=\dfrac{\mathrm{h}-8}{2} \Rightarrow \mathrm{h}=-16$
$5=\dfrac{\mathrm{k}+12}{2} \Rightarrow \mathrm{K}=-2$
$\therefore$ Equation of imaged circle is
$(x+16)^{2}+(y+2)^{2}=25$
$x^{2}+y^{2}+32 x+4 y+235=0$
1. The circle passing through the point $(-1,0)$ and touching the $y$-axis at $(0,2)$ also passes through the point
(a) $\left(\dfrac{-3}{2} 10\right)$
(b) $\left(\dfrac{-5}{2}, 12\right)$
(c) $\left(\dfrac{-3}{2}, \dfrac{5}{2}\right)$
(d) $(-4,0)$
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Solution:
Family of circles passing through a point $(0,2)$ and touching line $\mathrm{x}=0$ (y-axis) is
$(x-0)^{2}+(y-2)^{2}+\lambda x=0$
It passes through $(-1,0)$
$\therefore 1+4-\lambda=0$
$\therefore \lambda=5$
$\therefore$ equation of circle is
$\mathrm{x}^{2}+\mathrm{y}^{2}+5 \mathrm{x}-4 \mathrm{y}+4=0$
It also passes through $\mathrm{A}\left(-\mathrm{x} _{1}, 0\right)$
$\therefore \mathrm{x} _{1}^{2}-5 \mathrm{x} _{1}+4=0$
$\left(\mathrm{x} _{1}-4\right)\left(\mathrm{x} _{1}-1\right)=0$
$\mathrm{x} _{1}=4, \mathrm{x} _{1}=1$
$\therefore$ it also passes through $\mathrm{A}(-4,0)$
2. Two parallel chords of a circle of radius 2 are at a distance $\sqrt{3}+1$ apart. If the chords subtend at the centre, angles of $\dfrac{\pi}{\mathrm{k}}$ and $\dfrac{2 \pi}{\mathrm{k}}$, where $\mathrm{k}>0$ then the value of $[\mathrm{k}]$ is.
( $[\mathrm{k}]$ denotes the greatest integer)
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Solution:
Let $\dfrac{\pi}{2 \mathrm{k}}=\alpha=\dfrac{1}{2} \angle \mathrm{AOB}=\angle \mathrm{AOM}$
Then $\angle \mathrm{CON}=2 \alpha$
$\operatorname{In} _{\Delta} \mathrm{AOM}$
$\cos \alpha=\dfrac{\mathrm{X}}{2}$
In $\Delta \mathrm{CON}$
$\cos 2 \alpha=\dfrac{\sqrt{3}+1-\mathrm{x}}{2}$
$\cos 2 \alpha=2 \cos ^{2} \alpha-1$
$\cos 2 \alpha=2 \dfrac{x^{2}}{4}-1$
$\therefore \quad \dfrac{\sqrt{3}+1-\mathrm{x}}{2}=\dfrac{\mathrm{x}^{2}}{2}-1$
$\begin{aligned} & \sqrt{3}+1-x=x^{2}-2 \\ & x^{2}+x-\sqrt{3}-3=0 \\ & \therefore \quad x=\dfrac{-1 \pm \sqrt{1+4(3+\sqrt{3})}}{2} \\ & =\dfrac{-1 \pm \sqrt{13+4 \sqrt{3}}}{2} \\ & =\dfrac{-1 \pm(2 \sqrt{3}+1)}{2} \quad\left(\therefore 13+4 \sqrt{3}=(2 \sqrt{3}+1)^{2}\right) \\ & \quad x=\dfrac{-1+2 \sqrt{3}+1}{2} \\ & \quad x=\sqrt{3} \\ & \quad \cos \alpha=\dfrac{\sqrt{3}}{2}=\cos \dfrac{\pi}{6} \\ & \quad \alpha=\dfrac{\pi}{6} \\ & \quad \text { Required angle }=\dfrac{\pi}{k}=2 \alpha=\dfrac{\pi}{3} \end{aligned}$
3. Let $\mathrm{ABC}$ and $\mathrm{AB}^{\prime}$ be two non-congruent triangles with sides $\mathrm{AB}=4, \mathrm{AC}=\mathrm{AC}^{\prime}=2 \sqrt{2}$ and angle $\beta=30^{\circ}$. The absolute value of the difference between the areas of these triangles is
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Solution:
Draw circle through $\mathrm{AC} \mathrm{C}^{\prime}$ and $\mathrm{AB}$ intersect the circle at $\mathrm{P}$
$\begin{aligned} & \mathrm{In} _{\Delta} \mathrm{ABD} \\ & =\dfrac{\mathrm{AD}}{\mathrm{AB}}=\sin 30 \\ & \dfrac{\mathrm{AD}}{4}=\dfrac{1}{2} \\ & \therefore \mathrm{AD}=2=\mathrm{DC}=\mathrm{C}^{\prime} \mathrm{D} \end{aligned}$
Difference of areas of $\triangle \mathrm{ABC}$ and $\triangle \mathrm{ABC}^{\prime}$ is $\Delta \mathrm{ACC}^{\prime}$
$\therefore \operatorname{ar}\left(\Delta \mathrm{ACC}^{\prime}\right)=\dfrac{1}{2} \times 4 \times 2=4$ sq.u.
4. The centres of two circles $\mathrm{C} _{1}$ and $\mathrm{C} _{2}$ each of unit radius are at a distance of 6 units from each other. Let $\mathrm{P}$ be the mid-point of the line segment joining the centres of $\mathrm{C} _{1}$ and $\mathrm{C} _{2}$ and $\mathrm{C}$ be a circle touching circles $\mathrm{C} _{1}$ and $\mathrm{C} _{2}$ externally. If a common tangents to $\mathrm{C} _{1}$ and $\mathrm{C}$ passing through $\mathrm{P}$ is also a common targent to $\mathrm{C} _{2}$ and $\mathrm{C} _{1}$ then the radius of the circle $\mathrm{C}$ is
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Solution:
In $\Delta \mathrm{CPC} _{2}$
$\mathrm{CP}^{2}=\left(\mathrm{CC} _{2}\right)^{2}-\left(\mathrm{C} _{2} \mathrm{P}\right)^{2}$
$\mathrm{k}^{2}=(\mathrm{r}+1)^{2}-9$
$\mathrm{k}^{2}=\mathrm{r}^{2}+2 \mathrm{r}-8$____________________(1)
In $\triangle \mathrm{PQC} _{2}$
$\mathrm{PQ}{ }^{2}=3^{2}-1^{2}$
$=8$
$\therefore$ In ${ } _{\Delta} \mathrm{CPQ}$
$\mathrm{k}^{2}=\mathrm{r}^{2}+8$____________________(2)
From (1) & (2)
$\mathrm{F}^{2}+8=\mathrm{r}^{2}+2 \mathrm{r}-8$
$2 \mathrm{r}=16$
$\therefore \mathrm{r}=8$
5. A straight line through the vertex $\mathrm{P}$ of a triangle $\mathrm{PQR}$ intersects the side $\mathrm{QR}$ at the point $\mathrm{S}$ and the circumcircle of the triangle $\mathrm{PQR}$ at the point $\mathrm{T}$. If $\mathrm{S}$ is not the centre of the circumcircle then
(a) $\dfrac{1}{\mathrm{PS}}+\dfrac{1}{\mathrm{ST}}<\dfrac{2}{\sqrt{\mathrm{QS} . \mathrm{SR}}}$
(b) $\dfrac{1}{\mathrm{PS}}+\dfrac{1}{\mathrm{ST}}>\dfrac{2}{\sqrt{\mathrm{QS} . \mathrm{SR}}}$
(c) $\dfrac{1}{\mathrm{PS}}+\dfrac{1}{\mathrm{ST}}<\dfrac{4}{\mathrm{QR}}$
(d) $\dfrac{1}{\mathrm{PS}}+\dfrac{1}{\mathrm{ST}}>\dfrac{4}{\mathrm{QR}}$
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Solution:
Points $\mathrm{P}, \mathrm{Q}, \mathrm{T}, \mathrm{R}$ are concyclic
$\therefore$ PS.ST $=$ QS.SR
$\Rightarrow \dfrac{\mathrm{PS}+\mathrm{ST}}{2} \geq \sqrt{\mathrm{PS} . \mathrm{ST}}(\mathrm{AM} \geq \mathrm{GM})$
$\therefore \mathrm{PT} \geq 2 \sqrt{\mathrm{PS} . \mathrm{ST}}$
and $\dfrac{1}{\mathrm{PS}}+\dfrac{1}{\mathrm{ST}} \geq \dfrac{2}{\sqrt{\mathrm{PSST}}}=\dfrac{2}{\sqrt{\mathrm{QSSR}}}$
Also, $\dfrac{\mathrm{SQ}+\mathrm{SR}}{2} \geq \sqrt{\mathrm{SQ.SR}}$
$\Rightarrow \dfrac{\mathrm{QR}}{2} \geq \sqrt{\mathrm{SQ} . \mathrm{SR}}$
$\Rightarrow \dfrac{1}{\sqrt{\mathrm{SQ} \cdot S R}} \geq \dfrac{2}{\mathrm{QR}}$
$\Rightarrow \dfrac{2}{\sqrt{\mathrm{SQ} \cdot \mathrm{SR}}} \geq \dfrac{4}{\mathrm{QR}}$
$\dfrac{1}{\mathrm{PS}}+\dfrac{1}{\mathrm{ST}} \geq \dfrac{2}{\sqrt{\mathrm{QS} . \mathrm{SR}}} \geq \dfrac{4}{\mathrm{QR}}$
Answer: (d)
6. Let $\mathrm{ABCD}$ be a quadrilateral with area 18 , with side $\mathrm{AB}$ parallel to the side $\mathrm{CD}$ and $A B=2 C D$. Let $A D$ be perpendicular to $A B$ and $C D$. If a circle is drawn inside the quadrilateral $\mathrm{ABCD}$ touching all the sides then its radius is
(a) 3
(b) 2
(c) $3 / 2$
(d) 1 .
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Solution:
$\mathrm{ABCD}$ is a trapezium $(\because$ AB parallel to $\mathrm{CD})$
$\therefore \operatorname{ar}(\mathrm{ABCD})=\dfrac{1}{2} \times \mathrm{h}$ (sum of parallel sides)
$=\dfrac{1}{2} \times 2 \mathrm{r}(2 \mathrm{a}+\mathrm{a})$
$18=\mathrm{r} \times 3 \mathrm{a}$
ar $=6$
$\mathrm{CB}$ is a tangent to the circle
$\therefore$ equation of tangent is
$y=\dfrac{-2 r}{a}(x-2 a) \Rightarrow 2 r x+a y-4 a r=0$
It is a tangent to the circle $(x-r)^{2}+(y-r)^{2}=r^{2}$
$\therefore \mathrm{r}=\left|\dfrac{2 \mathrm{r}^{2}+\mathrm{ar}-4 \mathrm{ar}}{\sqrt{4 \mathrm{r}^{2}+\mathrm{a}^{2}}}\right|$
$r \sqrt{4 r^{2}+a^{2}}=2 r^{2}-3 a r$
$\sqrt{4 \mathrm{r}^{2}+\mathrm{a}^{2}}=2 \mathrm{r}-3 \mathrm{a}$
Squaring
$4 r^{2}+a^{2}=4 r^{2}+9 a^{2}-12 a r$
$12 \mathrm{r}=8 \mathrm{a} \Rightarrow 3 \mathrm{r}=2 \mathrm{a}$
ar $=6$
$\mathrm{r}=\dfrac{2 \mathrm{a}}{3}$
$\dfrac{2 \mathrm{a}^{2}}{3}=6$
$\mathrm{a}^{2}=9$
$\mathrm{a}= \pm 3$
$\therefore \mathrm{r}=2$
Answer: (b)
7. The radius of the least circle passing through the point $(8,4)$ and cutting the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=40$ orthogonally is
(a) $\sqrt{5}$
(b) $\sqrt{7}$
(c) $2 \sqrt{5}$
(d) $4 \sqrt{5}$
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Solution:
Let the circle be $x^{2}+y^{2}+2 g x+2 f y+c=0$_______________(1)
Given circle is $\mathrm{x}^{2}+\mathrm{y}^{2}=40$__________________________(2)
These two circles are orthogonal
$\therefore \mathrm{c}-40=0 \Rightarrow \mathrm{c}=40$
(1) passes through $(8,4)$
$64+16+16 \mathrm{~g}+8 \mathrm{f}+40=0$
$120+16 \mathrm{~g}+8 \mathrm{f}=0$
$\mathrm{f}+2 \mathrm{~g}+15=0$ or $\mathrm{f}=-(2 \mathrm{~g}+15)$
radius $=\sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}}$
$=\sqrt{g^{2}+(2 g+15)^{2}-40}$
For least circle radius must be minimum
Let $f(g)=g^{2}+(2 g+15)^{2}-40$ is minimum
$\mathrm{f}^{\prime}(\mathrm{g})=2 \mathrm{~g}+4(2 \mathrm{~g}+15)=0$
$10 \mathrm{~g}=-60$
$\mathrm{g}=-6$
$\mathrm{f}^{\prime \prime}(\mathrm{g})=10>0$ minimum
$\mathrm{f}=-(-12+15)=-3$
Equation of circle is $x^{2}+y^{2}-12 x-6 y+40=0$
radius $=\sqrt{36+9-40}$
$=\sqrt{5}$
Answer: (a)
8. $P$ is a point $(a, b)$ in the first quadrant. If the two circles which pass throngh $P$ and touch both the co-ordinate axes cut at right angles, then
(a) $\mathrm{a}^{2}-6 \mathrm{ab}+\mathrm{b}^{2}=0$
(b) $\mathrm{a}^{2}+2 \mathrm{ab}-\mathrm{b}^{2}=0$
(c) $\mathrm{a}^{2}-4 \mathrm{ab}+\mathrm{b}^{2}=0$
(d) $\mathrm{a}^{2}-8 \mathrm{ab}+\mathrm{b}^{2}=0$
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Solution:
Equation of the two circles be
$(\mathrm{x}-\mathrm{r})^{2}+(\mathrm{y}-\mathrm{r})^{2}=\mathrm{r}^{2} \Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{xr}-2 \mathrm{yr}+\mathrm{r}^{2}=0$
These two circles passes through $(\mathrm{a}, \mathrm{b})$
$\therefore(\mathrm{a}-\mathrm{r})^{2}+(\mathrm{b}-\mathrm{r})^{2}=\mathrm{r}^{2}$
$\mathrm{a}^{2}+\mathrm{r}^{2}-2 \mathrm{ar}+\mathrm{b}^{2}+\mathrm{r}^{2}-2 \mathrm{br}-\mathrm{r}^{2}=0$
$\mathrm{r}^{2}-2 \mathrm{r}(\mathrm{a}+\mathrm{b})+\left(\mathrm{a}^{2}+\mathrm{b}^{2}\right)=0$
It is a quadratic equation in $\mathrm{r}$
$\therefore \mathrm{r} _{1}+\mathrm{r} _{2}=2(\mathrm{a}+\mathrm{b})$ and $\mathrm{r} _{1} \mathrm{r} _{2}=\mathrm{a}^{2}+\mathrm{b}^{2}$
Condition for orthogonality is
$\begin{aligned} & 2 \mathrm{~g}_1 \mathrm{~g}_2+2 \mathrm{f}_1 \mathrm{f}_2=\mathrm{C}_1+\mathrm{C}_2 \\ & 2 \mathrm{r}_1 \mathrm{r}_2+2 \mathrm{r}_1 \mathrm{r}_2=\mathrm{r}_1{ }^2+\mathrm{r}_2{ }^2 \\ & 4 \mathrm{r}_1 \mathrm{r}_2=\mathrm{r}_1{ }^2+\mathrm{r}_2{ }^2 \\ & 6 \mathrm{r}_1 \mathrm{r}_2=\mathrm{r}_1{ }^2+\mathrm{r}_2{ }^2+2 \mathrm{r}_1 \mathrm{r}_2 \\ & 6 \mathrm{r}_1 \mathrm{r}_2=\left(\mathrm{r}_1+\mathrm{r}_2\right)^2 \\ & 6\left(\mathrm{a}^2+\mathrm{b}^2\right)=4(\mathrm{a}+\mathrm{b})^2 \\ & 6 \mathrm{a}^2+6 \mathrm{~b}^2=4 \mathrm{a}^2+4 \mathrm{~b}^2+8 \mathrm{ab} \\ & 2 \mathrm{a}^2+2 \mathrm{~b}^2-8 \mathrm{ab}=0 \\ & \mathrm{a}^2+\mathrm{b}^2-4 \mathrm{ab}=0 \end{aligned}$
Answer: (c)
9. A circle $S \equiv 0$ passes through the common points of family of circles $x^{2}+y^{2}+\lambda x-4 y+3=0(\lambda \in R)$ and have minimum area then
(a) area of $S \equiv 0$ is $\pi$ sq.u
(b) radius of director circle of $S \equiv 0$ is $\sqrt{2}$
(c) Radius of director circle of $\mathrm{S} \equiv 0$ for $\mathrm{x}$-axis is 1 unit
(d) $\mathrm{S} \equiv 0$ never cuts $|2 \mathrm{x}|=1$
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Solution:
$\left(\mathrm{x}^{2}+\mathrm{y}^{2}-4 \mathrm{y}+3\right)+\lambda \mathrm{x}=0$
$\therefore \mathrm{x}=0$ and $\mathrm{y}^{2}-4 \mathrm{y}+3=0$
$(\mathrm{y}-3)(\mathrm{y}-1)=0$
$\mathrm{y}=3,1$
$\therefore(0,3)(0,1)$ are common points.
$\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0$
passing through $(0,3) &(0,1)$
$9+6 \mathrm{f}+\mathrm{c}=0$________________________(1)
$1+2 \mathrm{f}+\mathrm{c}=0$________________________(2)
(1) $-(2)$ we get
$8+4 \mathrm{f}=0$
$\mathrm{f}=-2$ and $\mathrm{c}=3$
$\therefore \mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}-4 \mathrm{y}+3=0$
radius $=\sqrt{\mathrm{g}^{2}+4-3}=\sqrt{\mathrm{g}^{2}+1}$
for minimum area radius must be minimum
Since $g^{2}+1$ is positive so $g$ must be zero
$\therefore$ radius $=1$
Area $=\pi \mathrm{r}^{2}=\pi$ sq.u.
Radius of director circle is $\sqrt{2}$ times the radius of the given circle.
$\therefore$ Radius of director circle is $\sqrt{2}$
Answer: (b)
10. Area of part of circle $x^{2}+y^{2}-4 x-6 y+12=0$ above the line $4 x+7 y-29=0$ is $\Delta$, then $[\Delta]=$ [.] is greatest integer function.
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Solution
Since line $4 x+7 y-29=0$
passes through the centre $(2,3)$ of the circle
$\therefore \quad$ The line is a diameter of a circle with radius $\sqrt{4+9-12}=1$
$\therefore \quad$ area of semi circle is $=\dfrac{1}{2} \pi \mathrm{r}^{2}$
$\hspace {3 cm}\begin{aligned} & =\dfrac{1}{2} \pi \\ & =\dfrac{3.14}{2}=1.57 \end{aligned}$
Hence $[\Delta]=[1.57]=1$
EXERCISE:
1. The number of common tangents that can be drawn to the circles $x^{2}+y^{2}-4 x-6 y-3=0$ and
(a) 1
(b) 2
(c) 3
(d) 4
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Answer: c2. A variable chord is drawn through the origin to the circle $x^{2}+y^{2}-2 a x=0$. The locus of the centre of the circle drawn on this chord as diameter is
(a) $x^{2}+y^{2}+a x=0$
(b) $x^{2}+y^{2}+a y=0$
(c) $x^{2}+y^{2}-a x=0$
(d) $x^{2}+y^{2}-a y=0$
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Answer: c3. If $O$ is the origin and $\mathrm{OP}, \mathrm{OQ}$ are distinct tangents to the circle $\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0$, the circumcentre of the triangle OPQ is
(a) $(-\mathrm{g},-\mathrm{f})$
(b) $(\mathrm{g}, \mathrm{f})$
(c) $(-\mathrm{f},-\mathrm{g})$
(d) none of these
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Answer: d4. Equation of the normal to the circle $x^{2}+y^{2}-4 x+4 y-17=0$ which passes through $(1,1)$ is
(a) $3 x+2 y-5=0$
(b) $3 x+y-4=0$
(c) $3 x+2 y-2=0$
(d) $3 \mathrm{x}-\mathrm{y}-8=0$
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Answer: b5. The equation of the circle touching the lines $|y|=x$ at a distance $\sqrt{2}$ unit from the origin is
(a) $x^{2}+y^{2}-4 x+2=0$
(b) $x^{2}+y^{2}+4 x-2=0$
(c) $x^{2}+y^{2}+4 x+2=0$
(d) none of these
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Answer: a6. The shortest distance from the point $(2,-7)$ to the circle $x^{2}+y^{2}-14 x-10 y-151=0$ is
(a) 1
(b) 2
(c) 3
(d) 4
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Answer: b7. The equation of the image of the circle $(x-3) 2+(y-2) 2=1$ by the mirror $x+y=19$ is
(a) $(x-14)^{2}+(y-13)^{2}=1$
(b) $(x-15)^{2}+(y-14)^{2}=1$
(c) $(x-16)^{2}+(y-15)^{2}=1$
(d) $(x-17)^{2}+(y-16)^{2}=1$
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Answer: d8. If $P$ and $Q$ are two points on the circle $x^{2}+y^{2}-4 x-4 y-1$ which are farthest and nearest respectively from the point $(6,5)$, then
(a) $\mathrm{P}=\left(-\dfrac{22}{5}, 3\right)$
(b) $\mathrm{Q}=\left(\dfrac{22}{5}, \dfrac{19}{5}\right)$
(c) $\mathrm{P}=\left(\dfrac{14}{3},-\dfrac{11}{5}\right)$
(d) $\mathrm{Q}=\left(-\dfrac{14}{3},-4\right)$
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Answer: b9. A circle of the coaxial system with limiting points $(0,0)$ and $(1,0)$ is
(a) $x^{2}+y^{2}-2 x=0$
(b) $x^{2}+y^{2}-6 x+3=0$
(c) $x^{2}+y^{2}=1$
(d) $x^{2}+y^{2}-2 x+1=0$
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Answer: d10. If a variable circle touches externally two given circles, then the locus of the centre of the variable circle is
(a) a straight line
(b) a parabola
(c) an ellipse
(d) a hyperbola
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Answer: dPASSAGE
For each natural number $\mathrm{k}$, let $\mathrm{C} _{\mathrm{k}}$ denotes the circle with radius $\mathrm{k}$ units and centre at the origin. On the $\mathrm{C} _{\mathrm{k}}$, a particle moves $\mathrm{k}$ units in the counter clockwise direction. After completing its motion on $\mathrm{C} _{\mathrm{k}}$, the particle moves to $\mathrm{C} _{\mathrm{k}+\ell}$ in some well defined manner, where $\ell>0$. The motion of the particle continues in this manner.
On the basis of above information, answer the following questions:
11. Let $\ell=1$, the particle starts at $(1,0)$. If the particle crossing the positive direction of the $\mathrm{x}$-axis for the first time on the circle $\mathrm{C} _{\mathrm{n}}$, then $\mathrm{n}$ is equal to
(a) 3
(b) 5
(c) 7
(d) 8
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Answer: c12. If $\mathrm{k} \in \mathrm{N}$ and $\ell=1$, the particle starts $(-1,0)$ the particle cross $\mathrm{x}$-axis again at
(a) $(3,0)$
(b) $(1,0)$
(c) $(4,0)$
(d) $(2,0)$
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Answer: c13. If $\mathrm{k} \in \mathrm{N}$ and $\ell=1$, the particle moves in the radial direction from circle $\mathrm{C} _{\mathrm{k}}$ to $\mathrm{C} _{\mathrm{k}}+1$. If particle starts form the point $(-1,0)$, then
(a) it will cross the + ve $y$-axis at $(0,4)$
(b) it will cross the - ve $y$-axis at $(0,-4)$
(c) it will cross the $+v e y$-axis at $(0,5)$
(d) it will cross the - ve y-axis at $(0,-5)$
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Answer: c14. If $\mathrm{k} \in \mathrm{N}$ and $\ell=\mathrm{R}$, particle moves tangentially from the circle $\mathrm{C} _{\mathrm{k}}$ to $\mathrm{C} _{\mathrm{k}+1}$, such that the length of tangent is equal to $k$ units itself. If particle starts form the point $(1,0)$, then
(a) the particle will cross $\mathrm{x}$-axis again at $\mathrm{x}=3$
(b) the particle will cross $x$-axis again at $x=4$
(c) the particle will cross +ve $x$-axis again at $x=2 \sqrt{2}$
(d) the particle will cross +ve $x$-axis again at $x \in(2 \sqrt{2}, 4)$
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Answer: d15. Let the particle starts from the point $(2,0)$ and moves $\pi / 2$ units, on circle $\mathrm{C} _{2}$ in the counterclockwise direction, then itn moves on the circle $\mathrm{C} _{3}$ along the tangential path, let this straight line (tangential path traced by particle) intersect the circle $\mathrm{C} _{3}$ at the points. A and B tangents drawn at $\mathrm{A}$ and $\mathrm{B}$ intersect at
(a) $\left(\dfrac{9}{2 \sqrt{2}} ; \dfrac{9}{2 \sqrt{2}}\right)$
(b) $(9 \sqrt{2}, 9 \sqrt{2})$
(c) $(9,9)$
(d) $(\sqrt{2}, \sqrt{2})$
BETA
