Question

Find the equations of the circles which have the radius $\sqrt{13}$ and which touch the line $2x-3y+1=0$ at $(1,1)$

• A. $x^2+y^2+6x+4y=0$ or $x^2+y^2-2x-8y+4=0$
• B. $x^2+y^2-6x-4y=0$ or $x^2+y^2+2x+8y+4=0$
• C. $x^2+y^2-6x+4y=0$ or $x^2+y^2+2x-8y+4=0$
• D. $x^2+y^2+6x-4y=0$ or $x^2+y^2-2x+8y-4=0$

Answer 1

Answer: (C)

$x^2+y^2-6x+4y=0$ or $x^2+y^2+2x-8y+4=0$

Given that, the circle have the radius $\sqrt{13}$ and touches the line $2x-3y+1=0$ at $(1,1)$.
Point circle at $(1,1)$ is $(x-1{)}^2+(y-1{)}^2=0$
i.e $S:x^2+y^2-2x-2y+2=0$
$\therefore$ Equation of the circle is of the form $S+\lambda P=0$ where $P$ is $2x-3y+1=0$
$\Rightarrow (x^2+y^2-2x-2y+2)+\lambda (2x-3y+1)=0$
$\Rightarrow x^2+y^2+(2\lambda -2)x-(3\lambda +2)y+(\lambda +2)=0$
Radius of above circle is $\sqrt{13}$.
$\Rightarrow (\lambda -1{)}^2+{\left(\frac{3\lambda +2}{2}\right)}^2-(\lambda +2)=13$
$\Rightarrow 4(\lambda -1{)}^2+(3\lambda +2{)}^2-4(\lambda +2)=52$
$\Rightarrow 13{\lambda }^2=52$
$\Rightarrow \lambda =\pm 2$
$\therefore$ Required equations of circles are $x^2+y^2+2x-8y+4=0$ and $x^2+y^2-6x+4y=0$
Hence, option C.