# How do you write an equation for a circle with tangent to 3x - 4y + 3 = 0 at (-1,0) radius 7?

Sep 13, 2016

Thus, there are two circles satisfying the given cond.

$\left(1\right) : {\left(x + \frac{26}{5}\right)}^{2} + {\left(y - \frac{28}{5}\right)}^{2} = 49$.

$\left(2\right) : {\left(x - \frac{16}{5}\right)}^{2} + {\left(y + \frac{28}{5}\right)}^{2} = 49$.

#### Explanation:

Suppose that $C \left(h , k\right)$ is the Centre of the circle.

Since, the pt. of contact $P \left(- 1 , 0\right)$ is on the circle,

$C P = r a \mathrm{di} u s = 7$

$\therefore {\left(h + 1\right)}^{2} + {k}^{2} = 49. \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(1\right)$.

Also, line $C P \bot \text{ the tgt. : } 3 x - 4 y + 3 = 0$

As slope of tgt. is $\frac{3}{4} , \therefore \text{slope of } C P = - \frac{4}{3}$.

But, slope of $C P \text{ is } \frac{k}{h + 1} \therefore \frac{k}{h + 1} = - \frac{4}{3}$.

$\therefore \frac{k}{4} = \frac{h + 1}{-} 3 = \mu , \text{ say, so that, } k = 4 \mu , h + 1 = - 3 \mu$.

Sub.ing in $\left(1\right)$, we have, $9 {\mu}^{2} + 16 {\mu}^{2} = 49 \Rightarrow \mu = \pm \frac{7}{5}$

$C a s e 1 : \mu = + \frac{7}{5}$

$\therefore k = 4 \mu = \frac{28}{5} , h = - 1 - 3 \mu = - 1 - \frac{21}{5} = - \frac{26}{5}$

So, the Centre is$C \left(h , k\right) = C \left(- \frac{26}{5} , \frac{28}{5}\right) , r a \mathrm{di} u s = 7$. Hence,

the eqn. of the circle, is, ${\left(x + \frac{26}{5}\right)}^{2} + {\left(y - \frac{28}{5}\right)}^{2} = 49$.

$C a s e 2 : \mu = - \frac{7}{5}$

In this case, the Centre is$C \left(\frac{16}{5} , - \frac{28}{5}\right)$, & the eqn. of circle is

${\left(x - \frac{16}{5}\right)}^{2} + {\left(y + \frac{28}{5}\right)}^{2} = 49$.

Thus, there are two circles satisfying the given cond.

$\left(1\right) : {\left(x + \frac{26}{5}\right)}^{2} + {\left(y - \frac{28}{5}\right)}^{2} = 49$.

$\left(2\right) : {\left(x - \frac{16}{5}\right)}^{2} + {\left(y + \frac{28}{5}\right)}^{2} = 49$.

Enjoy Maths.!