The equation of the circle with centre $(2,3)$ and touching the line $3 x-4 y+1=0$ is

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Aicharya · Accepted Answer

Centre $c=(2,3)$ radius $=$ Perpendicular distance from centre $(2,3)$ to the line $3 x-4 y+1=0$ $\begin{aligned} & r=\frac{|3(2)-4(3)+1|}{\sqrt{3^2+(-4)^2}}=\frac{|7-12|}{\sqrt{25}} \ & r=\frac{5}{5}=1 \end{aligned}$ Equation of circle is $\begin{aligned} (x-2)^2+(y-3)^2 & =(1)^2 \ x^2+4-4 x+y^2+9-6 y & =1 \ x^2+y^2-4 x-6 y+13-1 & =0 \ x^2+y^2-4 x-6 y+12 & =0 \end{aligned}$ Hence, option (d) is correct.

The equation of the circle with centre \((2,3)\) and touching the line \(3 x-4 y+1=0\) is

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