Question

A circle $S$ passes through the point $(0,1)$ and is orthogonal to the circles $(x-1{)}^2+y^2=16$ and $x^2+y^2=1$. Then

• A. Radius of $S$ is $8$
• B. Radius of $S$ is $7$
• C. Centre of $S$ is $(-7,1)$
• D. Centre of $S$ is $(-8,1)$

Answer 1

Answer: (B), (C)

Centre of $S$ is $(-7,1)$

Let the equation of circle be
$x^2+y^2+2gx+2fy+c=0$
It passes through $(0,1)$
$\therefore 1+2f+c=0\dots \dots \left(i\right)$
This circle is orthogonal to $(x-1{)}^2+y^2=16$
i.e., $x^2+y^2-2x-15=0$
and $x^2+y^2-1=0$
$\therefore$ We should have
$2g(-1)+2f\left(0\right)=c-15$
or $2g+c-15=0\dots \dots \left(ii\right)$
and $2g\left(0\right)+2f\left(0\right)=c-1$
or $c=1\dots \dots \left(iii\right)$
Solving $\left(i\right),\left(ii\right)$ and $\left(iii\right)$, we get
$c=1,g=7,f=-1$
$\therefore$ Required circle is
$x^2+y^2+14x-2y+1=0$
With centre $(-7,1)$ and radius = $7$.