Circle Ques 62
- Find the coordinates of the point at which the circles $x^{2}-y^{2}-4 x-2 y+4=0$ and $x^{2}+y^{2}-12 x-8 y+36=0$ touch each other. Also, find equations of common tangents touching the circles the distinct points.
(1993, 5M)
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Answer:
Correct Answer: 62.$(x-5)^{2}+(y-5)^{2}=5^{2}$ and $(x+3)^{2}+(y+1)^{2}=5^{2}$
Solution:
Formula:
- Two circles touch each other externally, if $C_1 C_2=r_1+r_2$ and internally if $C_1 C_2=|r_1 - r_2|$
Given circles are $x^{2}+y^{2}-4 x-2 y+4=0$,
whose centre $C _1(2,1)$ and radius $r _1=1$
and $\quad x^{2}+y^{2}-12 x-8 y+36=0$
whose centre $C _2(6,4)$ and radius $r _2=4$
The distance between the centres is measured
$$ \sqrt{(6-2)^{2}+(4-1)^{2}}=\sqrt{16+9}=5 $$
$$ \Rightarrow \quad C _1 C _2=r _1 r _2 $$
Therefore, the circles touch each other externally, and the point of contact divides the line joining the two centres internally in the ratio of their radii, $1: 4$.
Therefore, $\quad x _1=\frac{1 \times 6+4 \times 2}{1+4}=\frac{14}{5}$
$$ y _1=\frac{1 \times 4+4 \times 1}{1+4}=\frac{8}{5} $$
Again, to determine the equations of common tangents touching the circles in distinct points, we know that, the tangents pass through a point which divides the line joining the two centres externally in the ratio of their radii, i.e. $r_1:r_2$.
Therefore, $x _2=\frac{1 \times 6-4 \times 2}{1-4}=\frac{-2}{-3}=\frac{2}{3}$
and $\quad y _2=\frac{1 \times 4-4 \times 1}{1-4}=0$
Now, let $m$ be the slope of the tangent, and this line passing through $(2/3, 0)$ is
$$ \begin{alignedat} y-0 & =m(x-\frac{2}{3}) \\ \Rightarrow \quad y-m x+\frac{2}{3} m & =0 \end{aligned} $$
This is tangent to the Ist circle, if perpendicular distance from centre equals radius.
$$ \begin{alignedat} & \therefore \quad \frac{1-2 m+(2 / 3) m}{\sqrt{1+m^{2}}}=1 \quad\left[\because C _1 \equiv(2,1) \text { and } r _1=1\right] \\ & \Rightarrow \quad 1-2 m+(2 / 3) m=\sqrt{1+m^{2}} \\ & \Rightarrow \quad 1-\frac{4}{3} m=\sqrt{1+m^{2}} \\ & \Rightarrow \quad 1+\frac{16}{9} m^{2}-\frac{8}{3} m=1+m^{2} \\ & \Rightarrow \quad \frac{7}{9} m^{2}-\frac{8}{3} m=0 \\ & \Rightarrow \quad \frac{7}{9} m - \frac{8}{3} = 0 \\ & \Rightarrow \quad m=0, m=\frac{24}{7} \end{aligned} $$
Hence, the equations of the two tangents are
$$ \begin{alignedat} & y=0 \text { and } y=\frac{24}{7} \quad x-\frac{2}{3} \\ \Rightarrow \quad & y=0 \text { and } 7 y-24 x+16=0 \end{aligned} $$
BETA
