Question

The equation of the circle having centre at (3, –4) and touching the line 5x + 12y – 12 = 0 is____________.

Answer 1

Circle with center (3, −4) has equation of the form (x − 3)² + (y + 4)² = r² ....(1)

Since circle touches the line 5x + 12y − 12

i.e. perpendicular distance of center to the line gives radius.

$$\begin{gathered} \mathrm{i}.\mathrm{e}.\mathrm{Distance}=\frac{\left|5\left(3\right)+12\left(-4\right)-12\right|}{\sqrt{5^2+{12}^2}}=\frac{\left(15-48-12\right)}{13} \\ =\frac{45}{13} \end{gathered}$$

$\mathrm{i}.\mathrm{e}.\mathrm{equation}\mathrm{of}\mathrm{circle}\mathrm{is}{\left(x-3\right)}^2+{\left(y+4\right)}^2={\left(\frac{45}{13}\right)}^2$