Answer :

The equation of a circle with centre $\left(h, k\right)$ and radius $r$ is given as

$\left(x-h\right)^{2}+\left(y-k\right)^{2}=r^{2}$

It is given that centre $\left(h, k\right)=\left(0,2\right)$ and radius $\left(r\right)=2$

Therefore, the equation of the circle is

$\left(x-0\right)^{2}+\left(y-2\right)^{2}=2^{2}$

$x^{2}+y^{2}+4-4 y=4$

$x^{2}+y^{2}-4 y=0$

Answer :

The equation of a circle with centre $\left(h, k\right)$ and radius $r$ is given as

$\left(x-h\right)^{2}+\left(y-k\right)^{2}=r^{2}$

It is given that centre $\left(h, k\right)=\left(-2,3\right)$ and radius $\left(r\right)=4$

Therefore, the equation of the circle is

$\left(x+2\right)^{2}+\left(y-3\right)^{2}=\left(4\right)^{2}$

$x^{2}+4 x+4+y^{2}-6 y+9=16$

$x^{2}+y^{2}+4 x-6 y-3=0$

Answer :

The equation of a circle with centre $\left(h, k\right)$ and radius $r$ is given as

$\left(x - h\right)^{2}+\left(y - k\right)^{2}=r^{2}$

It is given that centre $\left(h, k\right)=\left(\dfrac{1}{2}, \dfrac{1}{4}\right)$ and radius $\left(r\right)=\dfrac{1}{12}$

Therefore, the equation of the circle is

$ \begin{aligned} & \left(x-\dfrac{1}{2}\right)^{2}+\left(y-\dfrac{1}{4}\right)^{2}=\left(\dfrac{1}{12}\right)^{2} \\ \\ & x^{2}-x+\dfrac{1}{4}+y^{2}-\dfrac{y}{2}+\dfrac{1}{16}=\dfrac{1}{144} \\ \\ & x^{2}-x+\dfrac{1}{4}+y^{2}-\dfrac{y}{2}+\dfrac{1}{16}-\dfrac{1}{144}=0 \\ \\ & 144 x^{2}-144 x+36+144 y^{2}-72 y+9-1=0 \\ \\ & 144 x^{2}-144 x+144 y^{2}-72 y+44=0 \\ \\ & 36 x^{2}-36 x+36 y^{2}-18 y+11=0 \\ \\ & 36 x^{2}+36 y^{2}-36 x-18 y+11=0 \end{aligned} $

Answer :

The equation of a circle with centre $\left(h, k\right)$ and radius $r$ is given as $\left(x - h\right)^{2}+\left(y - k\right)^{2}=r^{2}$

It is given that centre $\left(h, k\right)=\left(1,1\right)$ and radius $\left(r\right)=\sqrt{2}$

Therefore, the equation of the circle is

$ \begin{aligned} & \left(x-1\right)^{2}+\left(y-1\right)^{2}=\left(\sqrt{2}\right)^{2} \\ \\ & x^{2}-2 x+1+y^{2}-2 y+1=2 \\ \\ & x^{2}+y^{2}-2 x-2 y=0 \end{aligned} $

Answer :

The equation of a circle with centre $\left(h, k\right)$ and radius $r$ is given as

$\left(x - h\right)^{2}+\left(y - k\right)^{2}=r^{2}$

It is given that centre $\left(h, k\right)=\left(-a, -b\right)$ and radius $\left(r\right)=\sqrt{a^{2}-b^{2}}$

Therefore, the equation of the circle is

$ \begin{aligned} & \left(x+a\right)^{2}+\left(y+b\right)^{2}=\left(\sqrt{a^{2}-b^{2}}\right)^{2} \\ \\ & x^{2}+2 a x+a^{2}+y^{2}+2 b y+b^{2}=a^{2}-b^{2} \\ \\ & x^{2}+y^{2}+2 a x+2 b y+2 b^{2}=0 \end{aligned} $

Answer :

The equation of the given circle is $\left(x+5\right)^{2}+\left(y-3\right)^{2}=36$

$\left(x+5\right)^{2}+\left(y-3\right)^{2}=36$

$ \lbrace x-\left(-5\right) \rbrace ^{2}+\left(y-3\right)^{2}=6^{2}$,

which is of the form $\left(x-h\right)^{2}+\left(y-k\right)^{2}=r^{2}$,

where $h=-5, k=3$, and $r=6$

Thus, the centre of the given circle is $\left(-5,3\right)$, while its radius is $6 .$

Answer :

The equation of the given circle is $ x^{2}+y^{2} - 4 x-8 y - 45=0$

$x^{2}+y^{2} -4 x - 8 y - 45=0$

$\Rightarrow\left(x^{2} - 4 x\right)+\left(y^{2} - 8 y\right)=45$

$\Rightarrow \lbrace x^{2} - 2\left(x\right)\left(2\right)+2^{2} \rbrace + \lbrace y^{2} - 2\left(y\right)\left(4\right)+4^{2} \rbrace $ $- 4 -16=45$

$\Rightarrow\left(x - 2\right)^{2}+\left(y - 4\right)^{2}=65$

$\Rightarrow\left(x - 2\right)^{2}+\left(y - 4\right)^{2}=\left(\sqrt{65}\right)^{2}$,

which is of the form $\left(x - h\right)^{2}+\left(y - k\right)^{2}=r^{2}$, where $h=2, k=4$, and $r=\sqrt{65}$

Thus, the centre of the given circle is $\left(2,4\right)$, while its radius is $\sqrt{65}$

Answer :

The equation of the given circle is $x^{2}+y^{2} - 8 x+10 y - 12=0$

$x^{2}+y^{2} - 8 x+10 y - 12=0$

$\Rightarrow\left(x^{2} - 8 x\right)+\left(y^{2}+10 y\right)=12$

$\Rightarrow \lbrace x^{2} - 2\left(x\right)\left(4\right)+4^{2} \rbrace + \lbrace y^{2}+2\left(y\right)\left(5\right)+5^{2} \rbrace - 16 - 25=12$

$\Rightarrow\left(x - 4\right)^{2}+\left(y+5\right)^{2}=53$

$\Rightarrow\left(x-4\right)^{2}+\left[ \lbrace y-\left(-5\right) \rbrace \right]^{2}=\left(\sqrt{53}\right)^{2}$,

which is of the form $\left(x - h\right)^{2}+\left(y - k\right)^{2}=r^{2}$, where $h=4, k=- 5,$ and $ \ \ r=\sqrt{53}$

Thus, the centre of the given circle is $\left( 4, - 5 \right)$, while its radius is $\sqrt{53}$

Answer :

The equation of the given circle is $2 x^{2}+2 y^{2} - x=0$

$ \begin{aligned} & 2 x^{2}+2 y^{2}-x=0 \\ \\ & \Rightarrow\left(2 x^{2}-x\right)+2 y^{2}=0 \\ \\ & \Rightarrow 2\left[\left(x^{2}-\dfrac{x}{2}\right)+y^{2}\right]=0 \\ \\ & \Rightarrow \left\lbrace x^{2}-2 \cdot x\left(\dfrac{1}{4}\right)+\left(\dfrac{1}{4}\right)^{2} \right\rbrace +y^{2}-\left(\dfrac{1}{4}\right)^{2}=0 \\ \\ & \Rightarrow\left(x-\dfrac{1}{4}\right)^{2}+\left(y-0\right)^{2}=\left(\dfrac{1}{4}\right)^{2}, \\ \\ &\text{ which is of the form }\left(x - h\right)^{2}+\left(y - k\right)^{2}=r^{2}, \text{ where } h=\dfrac{1}{4}, k=0, \text{ and } r=\dfrac{1}{4} \end{aligned} $

Thus, the centre of the given circle is $\left(\dfrac{1}{4}, 0\right)$, while its radius is $\dfrac{1}{4}$

Answer :

Let the equation of the required circle be $\left(x - h\right)^{2}+\left(y - k\right)^{2}=r^{2}$

Since the circle passes through points $\left(4,1\right)$ and $\left(6,5\right)$,

$\left(4 - h\right)^{2}+\left(1 - k\right)^{2}=r^{2}\qquad \ldots(1)$

$\left(6 - h\right)^{2}+\left(5 - k\right)^{2}=r^{2}\qquad \ldots(2)$

Since the centre $\left(h, k\right)$ of the circle lies on line $4 x+y=16$,

$4 h+k=16$

From equations $\left(1\right)$ and $\left(2\right),$ we obtain

$\left(4 - h\right)^{2}+\left(1 - k\right)^{2}=\left(6 - h\right)^{2}+\left(5 -{2} k\right)^{2}$

$\Rightarrow 16 - 8 h+h^{2}+1-2 k+k^{2}=36-12 h+h^{2}+25-10 k+k^{2}$

$\Rightarrow 16 - 8 h+1 - 2 k=36 - 12 h+25 - 10 k$

$\Rightarrow 4 h+8 k=44\qquad \ldots(3)$

$\Rightarrow h+2 k=11\qquad \ldots(4)$

On solving equations $\left(3\right)$ and $\left(4\right),$ we obtain $h=3$ and $k=4$

On substituting the values of $h$ and $k$ in equation $\left(1\right),$ we obtain

$\left(4 - 3\right)^{2}+\left(1 -4\right)^{2}=r^{2}$

$\Rightarrow\left(1\right)^{2}+\left(- 3\right)^{2}=r^{2}$

$\Rightarrow 1+9=r^{2}$

$\Rightarrow r^{2}=10$

$\Rightarrow r=\sqrt{10}$

Thus, the equation of the required circle is

$\left(x - 3\right)^{2}+\left(y - 4\right)^{2}=\left(\sqrt{10}\right)^{2}$

$x^{2} - 6 x+9+y^{2} - 8 y+16=10$

$x^{2}+y^{2} -6 x-8 y+15=0$

Answer :

Let the equation of the required circle be $\left(x - h\right)^{2}+\left(y - k\right)^{2}=r^{2}$

Since the circle passes through points $\left(2,3\right)$ and $\left( - 1,1 \right),$

$\left(2 - h\right)^{2}+\left(3 - k\right)^{2}=r^{2} \qquad\ldots\left(1\right)$

$\left(- 1 - h\right)^{2}+\left(1 - k\right)^{2}=r^{2}\qquad \ldots(2)$

Since the centre $\left(h, k\right)$ of the circle lies on line $ x -3 y - 11=0 $,

$h - 3 k=11\qquad \ldots(3)$

From equations $\left(1\right)$ and $\left(2\right),$ we obtain

$ \left(2-h\right)^2 + \left(3-k\right)^2 =\left(-1-h\right)^2 + \left(1-k\right)^2 $

$\Rightarrow 4 - 4 h+h^{2}+9 - 6 k+k^{2}=1+2 h+h^{2}+1 - 2 k+k^{2}$

$\Rightarrow 4 - 4 h+9 - 6 k=1+2 h+1 - 2 k$

$\Rightarrow 6 h+4 k=11\qquad \ldots(4)$

On solving equations $\left(3\right)$ and $\left(4\right),$ we obtain $h=\dfrac{7}{2}$ and $k=\dfrac{-5}{2}$

On substituting the values of $h$ and $k$ in equation $\left(1\right),$ we obtain

$\left(2-\dfrac{7}{2}\right)^{2}+\left(3+\dfrac{5}{2}\right)^{2}=r^{2}$

$\Rightarrow\left(\dfrac{4-7}{2}\right)^{2}+\left(\dfrac{6+5}{2}\right)^{2}=r^{2}$

$\Rightarrow\left(\dfrac{-3}{2}\right)^{2}+\left(\dfrac{11}{2}\right)^{2}=r^{2}$

$\Rightarrow \dfrac{9}{4}+\dfrac{121}{4}=r^{2}$

$\Rightarrow \dfrac{130}{4}=r^{2}$

Thus, the equation of the required circle is

$ \begin{aligned} & \left(x-\dfrac{7}{2}\right)^{2}+\left(y+\dfrac{5}{2}\right)^{2}=\dfrac{130}{4} \\ \\ & \left(\dfrac{2 x-7}{2}\right)^{2}+\left(\dfrac{2 y+5}{2}\right)^{2}=\dfrac{130}{4} \\ \\ & 4 x^{2}-28 x+49+4 y^{2}+20 y+25=130 \\ \\ & 4 x^{2}+4 y^{2}-28 x+20 y-56=0 \\ \\ & 4\left(x^{2}+y^{2}-7 x+5 y-14\right)=0 \\ \\ & x^{2}+y^{2}-7 x+5 y-14=0 \end{aligned} $

Answer :

Let the equation of the required circle be $\left(x - h\right)^{2}+\left(y - k\right)^{2}=r^{2}$

Since the radius of the circle is 5 and its centre lies on the $x$-axis, $k=0$ and $r=5$

Now, the equation of the circle becomes $\left(x - h\right)^{2}+y^{2}=25$

It is given that the circle passes through point $\left(2,3\right)$

$\therefore \ \left(2-h\right)^{2}+3^{2}=25$

$\Rightarrow\left(2-h\right)^{2}=25-9$

$\Rightarrow\left(2-h\right)^{2}=16$

$\Rightarrow 2-h= \pm \sqrt{16}= \pm 4$

If $2-h=4$, then $h=-2$

If $2-h=-4$, then $h=6$

When $h=-2$ , the equation of the circle becomes

$\left(x+2\right)^{2}+y^{2}=25$

$x^{2}+4 x+4+y^{2}=25$

$x^{2}+y^{2}+4 x - {21}=0$

When $h=6$, the equation of the circle becomes

$\left(x \text{ - 6 }\right)^{2}+y^{2}=25$

$ x^2 - 12x + 36 +y^2 = 25 $

$x^{2}+y^{2} - 12 x+11=0$

Answer :

Let the equation of the required circle be $\left(x - h\right)^{2}+\left(y - k\right)^{2}=r^{2}$

Since the circle passes through $\left(0,0\right)$,

$\left(0 - h\right)^{2}+\left(0 - k\right)^{2}=r^{2}$

$\Rightarrow h^{2}+k^{2}=r^{2}$

The equation of the circle now becomes $\left(x - h\right)^{2}+\left(y - k\right)^{2}=h^{2}+k^{2}$

It is given that the circle makes intercepts $a$ and $b$ on the coordinate axes.

This means that the circle passes through points $\left(a, 0\right)$ and $\left(0, b\right)$ Therefore,

$\left(a - h\right)^{2}+\left(0 - k\right)^{2}=h^{2}+k^{2}\qquad \ldots(1)$

$\left(0 - h\right)^{2}+\left(b - k\right)^{2}=h^{2}+k^{2}\qquad \ldots(2)$

From equation $\left(1\right),$ we obtain

$a^{2} - 2 a h+h^{2}+k^{2}=h^{2}+k^{2}$

$\Rightarrow a^{2} - 2 a h=0$

$\Rightarrow a\left(a - 2 h\right)=0$

$\Rightarrow a=0$ or $\left(a - 2 h\right)=0$

However, $a \neq 0$

Hence, $\left(a - 2 h\right)=0 \Rightarrow h=2$

From equation $\left(2\right),$ we obtain

$h^{2}+b^{2} - 2 b k+k^{2}=h^{2}+k^{2}$

$\Rightarrow b^{2} - 2 b k=0$

$\Rightarrow b\left(b - 2 k\right)=0$

$\Rightarrow b=0$ or $\left(b - 2 k\right)=0$

However, $b \neq 0$

Hence, $\left( b - 2 k\right)=0 \Rightarrow k=\dfrac{b}{2}$

Thus, the equation of the required circle is

$ \begin{aligned} & \left(x-\dfrac{a}{2}\right)^{2}+\left(y-\dfrac{b}{2}\right)^{2}=\left(\dfrac{a}{2}\right)^{2}+\left(\dfrac{b}{2}\right)^{2} \\ \\ & \Rightarrow\left(\dfrac{2 x-a}{2}\right)^{2}+\left(\dfrac{2 y-b}{2}\right)^{2}=\dfrac{a^{2}+b^{2}}{4} \\ \\ & \Rightarrow 4 x^{2}-4 a x+a^{2}+4 y^{2}-4 b y+b^{2}=a^{2}+b^{2} \\ \\ & \Rightarrow 4 x^{2}+4 y^{2}-4 a x-4 b y=0 \\ \\ & \Rightarrow x^{2}+y^{2}-a x-b y=0 \end{aligned} $

Answer :

The centre of the circle is given as $\left(h, k\right)=\left(2,2\right)$

Since the circle passes through point $\left(4,5\right)$, the radius $\left(r\right)$ of the circle is the distance between the points $\left(2,2\right)$ and $\left(4, 5\right).$

$ \therefore r=\sqrt{\left(2-4\right)^{2}+\left(2-5\right)^{2}}=\sqrt{\left(-2\right)^{2}+\left(-3\right)^{2}}=\sqrt{4+9}=\sqrt{13} $

Thus, the equation of the circle is

$ \begin{aligned} & \left(x-h\right)^{2}+\left(y-k\right)^{2}=r^{2} \\ \\ & \left(x-2\right)^{2}+\left(y-2\right)^{2}=\left(\sqrt{13}\right)^{2} \\ \\ & x^{2}-4 x+4+y^{2}-4 y+4=13 \\ \\ & x^{2}+y^{2}-4 x-4 y-5=0 \end{aligned} $

Answer :

The equation of the given circle is $x^{2}+y^{2}=25$

$x^{2}+y^{2}=25$

$\Rightarrow\left(x - 0\right)^{2}+\left(y - 0\right)^{2}=5^{2}$, which is of the form $\left(x - h\right)^{2}+\left(y - k\right)^{2}=r^{2}$,

where $h=0, k=0$, and $r=5$

$\therefore \ $ Centre $=\left(0,0\right)$ and radius $=5$

Distance between point $\left(-2.5, 3.5\right)$ and centre $\left(0,0\right)$

$ \begin{aligned} & =\sqrt{\left(-2.5-0\right)^{2}+\left(3.5-0\right)^{2}} \\ \\ & =\sqrt{6.25+12.25} \\ \\ & =\sqrt{18.5} \\ \\ & =4.3\left(\text{ approx. }\right)<5 \end{aligned} $

Since the distance between point $\left(-2.5,3.5\right)$ and centre $\left(0,0\right)$ of the circle is less than the radius of the circle, point $\left(-2.5, 3.5\right)$ lies inside the circle.