Question

The equation of the circle passing through $(2,3)$ and touching the lines $x-2=0,3x-4y+1=0$ is

• A. $x^2+y^{2}x-8y+3=0$, $4x^2+4y^2+22x-21y+62=0$
• B. $x^2+y^2-x-5y+1=0$, $x^2+y^2-x-6y+5=0$
• C. $x^2+y^2+x-6y+3=0$, $4x^2+4y^2-21x-24y+62=0$
• D. $4x^2+4y^2-23x-23y+63=0$, $4x^2+4y^2-21x-24y+62=0$

Answer 1

The correct option is C $x^2+y^2+x-6y+3=0$, $4x^2+4y^2-21x-24y+62=0$

Let $x^2+y^2+2gx+2fy+c=0$ ...... $\left(A\right)$ be the equation of the circle.
Hence, this circle passes through $(2,3)$ and touches the lines $x=2$ and $3x-4y+1=0$
The point $(2,3)$ lies on the circle. Hence,
$13+4g+6f+c=0$ ...(1)
The distance of the centre from the line $x=2$ must be equal to the radius of the circle.

Hence,
$\sqrt{g^2+f^2-c}=\left|\frac{-g-2}{1}\right|=|g+2|$
$f^2-c=4+4g$ .... (2)

Similarly, the distance of the centre from the other line should also be equal to the radius.

Hence,
$\therefore \sqrt{g^2+f^2-c}=\left|\frac{-3g+4f+1}{5}\right|$
$25g^2+25f^2-25c=9g^2+16f^2+1-24gf+8f-6g$ ...(3)

From (1) and (2), we get
$13+6f+f^2-4=0$
$\Rightarrow f^2+6f+9=0$
$\Rightarrow f=\frac{-6\pm \sqrt{36-36}}{2}$
$\Rightarrow f=-3$

Equation (3) becomes

$25(g^2-c)+25\times 9-16\times 9=9g^2+1+72g-24-6g$
$16g^2-25c+104-66g=0$ ...(4)

From $\left(1\right)$ we get

$4g+c-5=0$ ...... $\left(5\right)$ $[\because f=-3]$

Solving $\left(4\right)$ and $\left(5\right)$ simultaneously, we get

$16g^2+34g-21=0$

$\therefore g=\frac{-34\pm \sqrt{{34}^2-4\left(16\right)(-21)}}{2\left(16\right)}$

$=\frac{-34\pm \sqrt{2500}}{32}=\frac{-34\pm 50}{32}$

$\therefore g=\frac{1}{2},-\frac{21}{8}$
and $c=3,\frac{31}{2}$ ....... [Substituting value of $g$ in $\left(5\right)$]

Substitute $g=\frac{1}{2}$ and $c=3$ and $f=-3$ in $\left(A\right)$ we get

$x^2+y^2+x-6y+3=0$

Substitute $g=-\frac{21}{8}$ and $c=\frac{31}{2}$ and $f=-3$ again in $\left(A\right)$ we get

$4x^2+4y^2-21x-24y+62=0$

Hence, C is correct.