Question

Find the equation of circles which touch $2x-3y+1=0$ at $(1,1)$ and having radius $\sqrt{13}$

Answer 1

Let the equation of the circle.

$(x-h{)}^2+(y-k{)}^2=r^2$ ----- (1)

$(h,k)$ is the coordinate of center, radius $\sqrt{13}$ and it is touches by the line $2x-3y+1=0$ at $(1,1)$

By distance formula

$\sqrt{(h-1{)}^2-(k-1{)}^2}=\sqrt{13}$

$h^2=13-k^2+2k+2h-2$ ------- (2)

Now,

Tangent is always perpendicular to the radius

Slope of the line $2x-3y+1\left(m_1\right)=\frac{2}{3}$

Slope of the radius $m_2=-\frac{3}{2}$

Also the slope of of radius

$m_2=\frac{k-1}{h-1}=-\frac{3}{2}$

$3h+2k=-5$ ------- (3)

$h=\frac{2k-5}{3}$ ----- (4)

squaring on both side on equation (3)

$9h^2+4k^2=25$

From (2) and (4)

$5k^2-6k-104$

$(5k-26)(k+4)=0$

$k=-4,\frac{26}{5}$

Therefore,

$h=-\frac{13}{3},\frac{27}{15}$

Hence the point of center

$(-4,-\frac{13}{3}),(\frac{26}{5},\frac{27}{15})$

Hence the equation of circle are,

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

or

$(x-\frac{26}{5}{)}^2+(y-\frac{27}{5}{)}^2=(\sqrt{13}{)}^2$