Question

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

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

Answer 1

Answer: (A), (C)

Radius of S is 7

Let the circle be $x^2+y^2+2gx+2fy+c=0.$

It is orthogonal with $(x-1{)}^2+y^2=16$

$\therefore 2(-g+0)=-15+c\Rightarrow -2g=-15+c$

It is also orthogonal with $x^2+y^2=1$

$\therefore 0=-1+c\Rightarrow c=1$

$\therefore g=7$

This circle passes through (0, 1)

$\therefore 1+2f+1=0\Rightarrow f=-1$

$\therefore$ The circle is $x^2+y^2+14x-2y+1=0$

$(x+7{)}^2+(y-1{)}^2=49$

$\therefore$ centre : ( -7, 1) and radius : 7