Question

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

• A. $x^2+y^2-8x-12y+27=0$
• B. $x^2+y^2-8x-12y-27=0$
• C. $x^2-y^2-8x-12y+27=0$
• D. $x^2+y^2+8x+12y+27=0$

Answer 1

The correct option is A $x^2+y^2-8x-12y+27=0$

Point of contact of a line with slope'm' on a circle of radius 'a' is $(\frac{am}{\sqrt{1+m^2}},\frac{-a}{\sqrt{1+m^2}}).$

Let centre of circle be $(h,k)$

$\Rightarrow x-h=\frac{5\left(\frac{-3}{4}\right)}{\sqrt{1+\frac{9}{16}}}$

$\Rightarrow x-h=\frac{\frac{-15}{4}}{\sqrt{\frac{25}{16}}}$

$\Rightarrow x-h=\frac{-15}{4}\times \frac{5}{4}$

$\Rightarrow x-h=-3$

$\Rightarrow 1-h=-3.$

$\Rightarrow h=4.$

$y-k=\frac{5\times 4}{5}$

$\Rightarrow 2-k=-4.$

$\Rightarrow k=6.$

$\therefore$ Equation of circle

$\Rightarrow (x-4{)}^2+(y-6{)}^2=25.$

$\Rightarrow x^2+y^2+16+36-8x-12y=25.$

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