Question

If the circles $x^2+y^2-4x-6y-12=0$ and $5(x^2+y^2)-8x-14y-32=0$ touch each other then their point of contact is

• A. $(0,0)$
• B. $(1,1)$
• C. $(-1,-1)$
• D. $(0,-1)$

Answer 1

The correct option is C $(-1,-1)$

Equations of the circles are $x^2+y^2-4x-6y-12=0$ and $x^2+y^2-\frac{8}{5}x-\frac{14}{5}y-\frac{32}{5}=0$
Centres are $A(2,3),B(\frac{4}{5},\frac{7}{5})$

$r_1=\sqrt{4+9+12}=5$
$r_2=\sqrt{\frac{16}{25}+\frac{49}{25}+\frac{32}{5}}=3$

Point of contact is E.C.S (External Centre of Similitude), which divides $\overline{AB}$ in the ratio $r_1:r_2$ externally.

So, point of contact is $(\frac{4-6}{2},\frac{7-9}{2})=(-1,-1)$


Alternate Solution:
Equations of the circles are $x^2+y^2-4x-6y-12=0$ and $x^2+y^2-\frac{8}{5}x-\frac{14}{5}y-\frac{32}{5}=0$
Centres are $A(2,3),B(\frac{4}{5},\frac{7}{5})$
$r_1=\sqrt{4+9+12}=5$

$r_2=\sqrt{\frac{16}{25}+\frac{49}{25}+\frac{32}{5}}=3$

$AB=\sqrt{{(2-\frac{4}{5})}^2+{(3-\frac{7}{5})}^2}=\sqrt{\frac{36}{25}+\frac{64}{25}}=\sqrt{\frac{100}{25}}=2$

$AB=r_1-r_2$ the circles touch internally.
Equation of common tangent of two circle touching each other, is
$S_2-S_1=0$
$\frac{12x}{5}+\frac{16y}{5}+\frac{28}{5}=0$
$\Rightarrow 3x+4y+7=0\cdots \left(i\right)$
Equation of line passing through their centres
$y-3=\frac{4}{3}(x-2)$
$\Rightarrow 3y-4x-1=0\cdots \left(ii\right)$
From equation $\left(i\right)$ and $\left(ii\right)$
$x=-1$ and $y=-1$
So, the point of contact is $(-1,-1)$