Question

Find the equation of the circle having radius 5 and which touches line 3x +4y - 11= 0 at point (1,2).

Answer 1

Let the equation of circle be-

$x^2+y^2+2gx+2fy+c=0$

Centre of circle $=(-g,-f)$

Radius of circle $=\sqrt{g^2+f^2-c}=5\left(\text{Given}\right)$

Distance between the centre of circle and the line $3x+4y-11=0$ is radius of circle, i.e., $5$.

Therefore,

$\frac{|-3g-4f-11|}{\sqrt{3^2+4^2}}=5$

$\Rightarrow 3g-4f-11=25\left(\text{Taking positive}\right)$

$\Rightarrow -3g-4f=36.....\left(1\right)$

Now,

Slope of normal to the circle at $(1,2)$-

$m=\frac{2+f}{1+g}$

Slope of given line $\left(m^{\prime }\right)=\frac{-3}{4}$

Therefore,

$m{m}^{\prime }=-1(\because \text{Normal and tangent to the circle at a point are always perpendicular})$

$\Rightarrow \frac{2+f}{1+g}=\frac{4}{3}$

$\Rightarrow 6+3f=4+4g$

$\Rightarrow 4g-3f-2=0.....\left(2\right)$

Multiplying equation $\left(1\right)$ and $\left(2\right)$ by $4$ and $3$ respectively.

$-12g-16f=144.....\left(3\right)$

$12g-9f=6.....\left(4\right)$

Adding equation $\left(3\right)\&\left(4\right)$, we have

$-25f=150$

$\Rightarrow f=-\frac{150}{25}=-6$

Substituting the value of $f$ in equation $\left(1\right)$, we have

$-3g+24=36$

$\Rightarrow g=\frac{36-24}{-3}=-4$

Thus the centre of circle is $(4,6)$.

Now,

$\because \sqrt{g^2+f^2-c}=5$

$\therefore \sqrt{16+36-c}=5$

Squaring both sides, we have

$52-c=25$

$\Rightarrow c=27$

Hence the equation of circle is-

$x^2+y^2-8x-12y+27=0$.