Question

The equation of a circle of radius 5 which lies within the circle $x^2+y^2+14x+10y-26=0$ and touches it at the point (-1, 3) is

• A. $x^2+y^2+8x+2y+8=0$
• B. $x^2+y^2+8x+2y-8=0$
• C. $x^2+y^2+8x+2y-14=0$
• D. None of the above

Answer 1

Answer: (B)

$x^2+y^2+8x+2y-8=0$

$x^2+y^2+14x+10y-26=0r=\sqrt{(7{)}^2+{\left(5\right)}^2-(-26)}=10$

Other circle lies in the given circle and its radius is $5$ , so it also passes through the centre of the given circle

Centre of given circle is $(-7,-5)$

Now we have the ends point of diameter of the other circle , so its centre is

$(\frac{-7-1}{2},\frac{-5+3}{2})(-4,-1)$

$r=5$

So the equation of other circle is

${(x+4)}^2+{(y+1)}^2=5^2{x}^2+y^2+8x+2y-8=0$

So option $B$ is correct.