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. centre of $S$ is $(-7,1)$
• B. centre of $S$ is $(-8,1)$
• C. radius of $S$ is $7$
• D. radius of $S$ is $8$

Answer 1

Answer: (A), (C)

radius of $S$ is $7$

Let the circle be,
$x^2+y^2+2gx+2fy+c=0$
Condition for orthogonality,
$2g_1{g}_2+2f_1{f}_2=c_1+c_2$
Orthogonal with
$(x-1{)}^2+y^2=16x^2+y^2-2x-15=0$
Applying condition for orthogonality,
$2g(-1)+0=c-15\cdots \left(1\right)$

Orthogonal with
$x^2+y^2-1=0$
Applying condition for orthogonality,
$0+0=c-1\Rightarrow c=1\cdots \left(2\right)$
From using equation $\left(1\right),$
$g=7$
Putting the point $(0,1)$ in the circle,
$1+2f+1=0\Rightarrow f=-1$
Therefore the equation of the circle will be,
$x^2+y^2+14x-2y+1=0\Rightarrow (x+7{)}^2+(y-1{)}^2=49$
Centre $=(-7,1)$
radius $=7$