Question

Equation of the circle whose radius is $5$ and which touches externally, then circle $x^2+y^2-2x-4y-20=0$ at the point $(5,5)$ is

• A. ${(x-9)}^2+{(y-6)}^2=5^2$
• B. ${(x-9)}^2+{(y-8)}^2=5^2$
• C. ${(x-7)}^2+{(y-3)}^2=5^2$
• D. none of these

Answer 1

The correct option is B ${(x-9)}^2+{(y-8)}^2=5^2$

Center of the given circle is $A(1,2)$and its radius is $\sqrt{1+{\left(2\right)}^2+20}=5$

point of contact $P$ is $(5,5)$

Let $B(h,k)$ be the center of the required circle of radius $5$

then $P$ is the mid point of $AB$ so that $\frac{h+1}{2}=5$ and $\frac{k+2}{2}=5$
$\Rightarrow h=9,k=8$
and an equation of the required circle is
$x^2+y^2-18x-16y+120=0$

$\Rightarrow {(x-9)}^2+{(y-8)}^2=5^2$