Answer :

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

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

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

Therefore, the equation of the circle is

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

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

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

Answer :

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

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

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

Therefore, the equation of the circle is

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

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

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

Answer :

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

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

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

Therefore, the equation of the circle is

$ \begin{aligned} & (x-\dfrac{1}{2})^{2}+(y-\dfrac{1}{4})^{2}=(\dfrac{1}{12})^{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 $(h, k)$ and radius $r$ is given as $(x - h)^{2}+(y - k)^{2}=r^{2}$

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

Therefore, the equation of the circle is

$ \begin{aligned} & (x-1)^{2}+(y-1)^{2}=(\sqrt{2})^{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 $(h, k)$ and radius $r$ is given as

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

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

Therefore, the equation of the circle is

$ \begin{aligned} & (x+a)^{2}+(y+b)^{2}=(\sqrt{a^{2}-b^{2}})^{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 $(x+5)^{2}+(y-3)^{2}=36$.

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

$\Rightarrow\{x-(-5)\}^{2}+(y-3)^{2}=6^{2}$, which is of the form $(x-h)^{2}+(y-k)^{2}=r^{2}$, where $h=-5, k=3$, and $r=6$.

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

Answer :

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

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

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

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

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

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

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

Answer :

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

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

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

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

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

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

and $r=\sqrt{53}$.

Thus, the centre of the given circle is $( 4, - 5 )$, 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(2 x^{2}-x)+2 y^{2}=0 \\ & \Rightarrow 2[(x^{2}-\dfrac{x}{2})+y^{2}]=0 \\ & \Rightarrow\{x^{2}-2 \cdot x(\dfrac{1}{4})+(\dfrac{1}{4})^{2}\}+y^{2}-(\dfrac{1}{4})^{2}=0 \\ & \Rightarrow(x-\dfrac{1}{4})^{2}+(y-0)^{2}=(\dfrac{1}{4})^{2}, \text{ which is of the form }(x - h)^{2}+(y - k)^{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 $(\dfrac{1}{4}, 0)$, while its radius is $\dfrac{1}{4}$.

Answer :

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

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

(4 - $h)^{2}+(1 \text{ - } k)^{2}=r^{2}$.

(6 - $h)^{2}+(5 - k)^{2}=r^{2}$

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

$4 h+k=16$

From equations (1) and (2), we obtain

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

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

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

$\Rightarrow 4 h+8 k=44$

$\Rightarrow h+2 k=11$.

On solving equations (3) and (4), we obtain $h=3$ and $k=4$.

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

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

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

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

$\Rightarrow r^{2}=10$

$\Rightarrow r=\sqrt{10}$

Thus, the equation of the required circle is

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

$x^{2}$ - $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 $(x - h)^{2}+(y - k)^{2}=r^{2}$.

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

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

$(- 1 - h)^{2}+(1 - k)^{2}=r^{2}$..

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

$h$ -$3 k=11$

From equations (1) and (2), we obtain

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

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

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

$\Rightarrow 6 h+4 k=11$..

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

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

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

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

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

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

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

Thus, the equation of the required circle is

$ \begin{aligned} & (x-\dfrac{7}{2})^{2}+(y+\dfrac{5}{2})^{2}=\dfrac{130}{4} \\ & (\dfrac{2 x-7}{2})^{2}+(\dfrac{2 y+5}{2})^{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(x^{2}+y^{2}-7 x+5 y-14)=0 \\ & x^{2}+y^{2}-7 x+5 y-14=0 \end{aligned} $

Answer :

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

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 $(x - h)^{2}+y^{2}=25$.

It is given that the circle passes through point $(2,3)$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer :

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

Since the circle passes through $(0,0)$,

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

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

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

It is given that the circle makes intercepts $a$ and $b$ on the coordinate axes. This means that the circle passes through points $(a, 0)$ and $(0, b)$. Therefore,

$(a - h)^{2}+(0 - k)^{2}=h^{2}+k^{2}$.

$(0 - h)^{2}+(b - k)^{2}=h^{2}+k^{2}$.

From equation (1), we obtain

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

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

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

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

However, $a \neq 0$; hence, $(a - 2 h)=0 \Rightarrow h=2$.

From equation (2), we obtain

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

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

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

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

However, $b \neq 0$; hence, ( $b$ - $2 k)=0 \Rightarrow k=\dfrac{b}{2}$.

Thus, the equation of the required circle is

$ \begin{aligned} & (x-\dfrac{a}{2})^{2}+(y-\dfrac{b}{2})^{2}=(\dfrac{a}{2})^{2}+(\dfrac{b}{2})^{2} \\ & \Rightarrow(\dfrac{2 x-a}{2})^{2}+(\dfrac{2 y-b}{2})^{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 $(h, k)=(2,2)$.

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

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

Thus, the equation of the circle is

$ \begin{aligned} & (x-h)^{2}+(y-k)^{2}=r^{2} \\ & (x-2)^{2}+(y-2)^{2}=(\sqrt{13})^{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 $x^{2}+y^{2}=25$.

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

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

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

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

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

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