Question

Equation of circle touching the line $x+y=4$ at $(1,3)$ and intersecting the circle $x^2+y^2=4$ orthogonally is

• A. $x^2+y^2-x+2y-15=0$
• B. $x^2+y^2-x-y-6=0$
• C. ${2x}^2+2y^2-x+y-22=0$
• D. ${2x}^2+2y^2-x-9y+8=0$

Answer 1

The correct option is D ${2x}^2+2y^2-x-9y+8=0$

The general equation of circle touching the line lx + my + n = 0 at given point $(x_1,y_1)$ is $(x-x_1{)}^2+(y-y_1{)}^2+\lambda (lx+my+n)=0$ where $\lambda \in R$

For two circles with equation,

$x^2+y^2+2g_{1}x+2f_{1}y+c_1=0$ and $x^2+y^2+2g_{2}x+2f_{2}y+c_2=0$,

Condition of orthogonality $\to 2g_1{g}_2+2f_1{f}_2=c_1+c_2$

So, for given condition equation will be,

$\to (x-1{)}^2+(y-3{)}^2+\lambda (x+y-4)=0$

$\to x^2+y^2+(\lambda -2)x+(\lambda -6)y+(10-4\lambda )=0$

Using condition of orthoganility,

$2(2-\lambda )\left(0\right)+2(6-\lambda )\left(0\right)=-4+10-4\lambda$

$\to \lambda =\frac{3}{2}$

So, Equation of circle,

$\to x^2+y^2-\frac{1}{2}x-\frac{9}{2}y+4=0$

$\to 2x^2+2y^2-x-9y+8=0$

Hence, (D) is the correct answer.