Question

The equation of the circle that intersects the circle $x^2+y^2+14x+6y+2=0$ orthogonally and whose centre is $(0,2)$

• A. $x^2+y^2-4y-6=0$
• B. $x^2+y^2+4y-14=0$
• C. $x^2+y^2+4y+14=0$
• D. $x^2+y^2-4y-14=0$

Answer 1

The correct option is D $x^2+y^2-4y-14=0$

The general equation of circle with center (h,k) and radius r is,

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

For two circles with equation

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

to be orthogonal,

$2(g_1{g}_2+f_1{f}_2)=c_1+c_2$

Now,

for the required circle equation is of form,

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

$x^2+y^2-4y+4-r^2=0$

Applying condition of orthogonality we get,

$2(3\times -2)=2+4-r^2$

$r^2=18$

So, equation of required circle.

$\to x^2+y^2-4y-14=0$

Hence, (D) is the correct answer.