# Find the equation of circle with radius 4 units and whose centre lies on the line 13x+4y=32 and which touches the line 3x+4y+28=0.?

Nov 26, 2017

see explanation.

#### Explanation:

Let equation of $L 1$ be $13 x + 4 y = 32$, and equation of $L 2$ be $3 x + 4 y + 28 = 0$.
Let center of the circle be $\left(h , k\right)$, as shown in the figure.
given that $\left(h , k\right)$ lies on $L 1$,
Substituting $h = x , y = k$ into $13 x + 4 y = 32$, we get,
$13 h + 4 k = 32$,
$\implies k = \frac{32 - 13 h}{4}$
recall that the perpendicular distance from a point $P \left(h , k\right)$ to a line $a x + b y + c = 0$ is $d = | a h + b k + c \frac{|}{\sqrt{{a}^{2} + {b}^{2}}}$,
given that $r = 4$, and the circle touches $L 2$,
$\implies 4 = | 3 h + 4 \times \frac{32 - 13 h}{4} + 28 \frac{|}{\sqrt{{3}^{2} + {4}^{2}}}$
$\implies 4 = | 3 h + 32 - 13 h + 28 \frac{|}{5}$
$\implies 20 = | - 10 h + 60 |$
$\implies h = 4 \mathmr{and} 8$
when $h = 4 , k = \frac{32 - 13 \times 4}{4} = - 5$
when $h = 8 , k = \frac{32 - 13 \times 8}{4} = - 18$
$\implies \left(h , k\right) = \left(4 , - 5\right) , \left(8 , - 18\right)$
equations of the circles are:
${\left(x - 4\right)}^{2} + {\left(y + 5\right)}^{2} = 16$,
and ${\left(x - 8\right)}^{2} + {\left(y + 18\right)}^{2} = 16$