Question

The centre of the circle touching the circles $x^2+y^2-4x-6y-12=0$, $x^2+y^2+6x+18y+26=0$ at their point of contact and passing through the point (1,-1) is

• A. $(1/3, -1)$
• B. $(1/5, 6/5)$
• C. $(1/2, 1/2)$
• D. $(-1/4, -1/2)$

Hint:

The equation of any circle passing through the intersection of two circles $S_1=0$ and $S_2=0$ is given by $S_1 + \lambda S_2 = 0$. This also applies to circles that are tangent, where the "intersection" is the single point of contact.

Answer 1

The Correct Option is A

Let the two given circles be $S_1=0$ and $S_2=0$.
$S_1: x^2+y^2-4x-6y-12=0$. Centre $C_1(2,3)$, Radius $r_1 = \sqrt{2^2+3^2-(-12)} = \sqrt{4+9+12}=\sqrt{25}=5$.
$S_2: x^2+y^2+6x+18y+26=0$. Centre $C_2(-3,-9)$, Radius $r_2 = \sqrt{(-3)^2+(-9)^2-26} = \sqrt{9+81-26}=\sqrt{64}=8$.
Distance between centers: $C_1C_2 = \sqrt{(-3-2)^2+(-9-3)^2} = \sqrt{(-5)^2+(-12)^2} = \sqrt{25+144} = \sqrt{169}=13$.
Sum of radii: $r_1+r_2=5+8=13$. Since $C_1C_2=r_1+r_2$, the two circles touch each other externally.
Any circle touching $S_1$ and $S_2$ at their point of contact belongs to the family of circles $S_1 + \lambda S_2 = 0$ for $\lambda \neq -1$. The case $\lambda=-1$ represents the common tangent at the point of contact.
The equation of the required circle is $(x^2+y^2-4x-6y-12) + \lambda(x^2+y^2+6x+18y+26) = 0$.
This circle passes through the point P(1,-1). Substitute $x=1, y=-1$ to find $\lambda$.
$(1^2+(-1)^2-4(1)-6(-1)-12) + \lambda(1^2+(-1)^2+6(1)+18(-1)+26) = 0$.
$(1+1-4+6-12) + \lambda(1+1+6-18+26) = 0$.
$(-8) + \lambda(16) = 0 \implies 16\lambda = 8 \implies \lambda = 1/2$.
The equation of the required circle is:
$(x^2+y^2-4x-6y-12) + \frac{1}{2}(x^2+y^2+6x+18y+26) = 0$.
$2(x^2+y^2-4x-6y-12) + (x^2+y^2+6x+18y+26) = 0$.
$2x^2+2y^2-8x-12y-24 + x^2+y^2+6x+18y+26 = 0$.
$3x^2+3y^2-2x+6y+2 = 0$.
Divide by 3: $x^2+y^2-\frac{2}{3}x+2y+\frac{2}{3}=0$.
The centre of this circle is $(-g, -f) = (-\frac{1}{2}(-\frac{2}{3}), -\frac{1}{2}(2)) = (\frac{1}{3}, -1)$.