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)

Given circles
$x^2+y^2-2x-15=0$
$x^2+y^2-1=0$
Radical axis $x+7=0$
Centre of circle lies on (1)
Let the centre be $(-7,k)$
Let equation be $x^2+y^2+14x-2ky+c=0$
Orthogonallity gives
$-14=c-15$
$\Rightarrow c=1...(2)$
$(0,1)\to 1-2k+1=0$
$\Rightarrow k=1$
Hence radius $=\sqrt{7^2+k^2-c}$
$=\sqrt{49+1-1}=7$
Alternate solution
Given circles $x^2+y^2-2x-15=0$
$x^2+y^2-1=0$
Let equation of circle $x^2+y^2+2gx+2fy+c=0$
Circle passes through $(0,1)$
$\Rightarrow 1+2f+c=0$
Applying condition of orthogonality
$-2g=c-15,0=c-1$
$\Rightarrow c=1,g=7,f=-1$
$r=\sqrt{49+1-1}=7;$
centre $(-7,1)$