Question

If two circles with centres at $(a,0)$ and $(-a,0)$ having radii $b$ and $c$ units respectively such that $a>b>c$. Then the point of contacts of common tangents to these two circles will always lie on

• A. $x^2+y^2=a^2\pm bc$
• B. $x^2-y^2=a^2\pm bc$
• C. $x^2+2y^2=a^2\pm bc$
• D. $2x^2+y^2=a^2\pm bc$

Answer 1

The correct option is A $x^2+y^2=a^2\pm bc$

The equation of the given circles be
$C_1:x^2+y^2-2ax+a^2-b^2=0C_2:x^2+y^2+2ax+a^2-c^2=0$
Let $P(h,k)$ be a point of contact of common tangents to these circles such that it lies on $C_1$. Then
$h^2+k^2-2ah+a^2-b^2=0\cdots \left(1\right)$
Now equation of tangent at $P$ on $C_1$ will be
$hx+ky-a(x+h)+a^2-b^2=0\Rightarrow x(h-a)+ky-ah+a^2-b^2=0$
This will also touch the circle $C_2$
$\therefore \frac{|-a(h-a)-ah+a^2-b^2|}{\sqrt{(h-a{)}^2+k^2}}=c\Rightarrow \frac{|2a^2-2ah-b^2|}{b}=c\Rightarrow 2ah-(2a^2-b^2)=\pm bc\Rightarrow 2ah-(a^2-b^2)=a^2\pm bc\Rightarrow h^2+k^2=a^2\pm bc\left(\text{using (1)}\right)$
Hence locus will be
$x^2+y^2=a^2\pm bc$