Question

Show that the circles:
$x^2+y^2-6x-9y+13=0$,
$x^2+y^2-2x-16y=0$
touch each other. Find the contact and the equation of common tangent at their point of contact.

Answer 1

Equation of first circle-

$S_1:x^2+y^2-6x-9y+13=0$

${(x-3)}^2+{(y-\frac{9}{2})}^2-9-\frac{81}{4}+13=0$

${(x-3)}^2+{(y-\frac{9}{2})}^2=\frac{65}{4}$

Here,

$r_1=\frac{\sqrt{65}}{2}$

$C_1=(3,\frac{9}{2})$

Equation of another circle-

$S_2:x^2+y^2-2x-16y=0$

${(x-1)}^2+{(y-8)}^2-1-64=0$

${(x-1)}^2+{(y-8)}^2=65$

Here,

$r_2=\sqrt{65}$

$C_2=(1,8)$

Distance between the centre of two circles-

$C_1{C}_2=\sqrt{{(3-1)}^2+{(8-\frac{9}{2})}^2}$

$C_1{C}_2=\sqrt{4+\frac{49}{4}}=\frac{\sqrt{65}}{2}$

$|r_2-r_1|=|\sqrt{65}-\frac{\sqrt{65}}{2}|=\frac{\sqrt{65}}{2}$

$\because C_1{C}_2=|r_1-r_2|$

Thus the two circles touches each other internally.

Since the circle touches each other internally. The point of contact $P$ divides $C_1{C}_2$ externally in the ratio $r_1:r_2$, i.e., $\frac{\sqrt{65}}{2}:\sqrt{65}=1:2$

Therefore, coordinates of $P$ are-

$(\frac{1\left(1\right)-2\left(3\right)}{1-2},\frac{1\left(8\right)-2\left(\frac{9}{2}\right)}{1-2})=(5,1)$

Therefore,

Equation of common tangent is-

$S_1-S_2=0$

$(5x+y-6\left(\frac{x+5}{2}\right)-9\left(\frac{y+1}{2}\right)+13)-(5x+y-2\left(\frac{x+5}{2}\right)-16\left(\frac{y+1}{2}\right))=0$

$\frac{-6x-9y-13}{2}+x+8y+13=0$

$4x-7y-13=0$

Hence the point of contact is $(5,1)$ and the equation of common tangent is $4x-7y-13=0$.