# Given the centre (0,0) and a tangent that has the equation 3x-4y=30, what is the equation of the circle?

Jul 17, 2016

${x}^{2} + {y}^{2} = 36$

#### Explanation:

Length of a perpendicular from a point $\left({x}_{1} , {y}_{1}\right)$ to the line $a x + b y + c = 0$ is given by $\frac{| a {x}_{1} + b {y}_{1} + c |}{\sqrt{{a}^{2} + {b}^{2}}}$

Hence length of perpendicular from $\left(0 , 0\right)$ to $3 x - 4 y - 30 = 0$ is $\frac{| 3 \cdot 0 + \left(- 4\right) \cdot 0 - 30 |}{\sqrt{{3}^{2} + {\left(- 4\right)}^{2}}} = \frac{30}{\sqrt{9 + 16}} = \frac{30}{5} = 6$

But this is radius of the circle. Hence, equation of circle will be

${\left(x - 0\right)}^{2} + {\left(y - 0\right)}^{2} = {6}^{2}$ or

${x}^{2} + {y}^{2} = 36$

Jul 17, 2016

Reqd. eqn. of the circle is ${x}^{2} + {y}^{2} = 36$.

#### Explanation:

To determine the eqn. of a circle, we need its (i) centre, and, (ii) the

radius. As the centre is given to be the Origin $O \left(0 , 0\right)$, let us

suppose that the radius of the circle is $r$.

Now the line $l : 3 x - 4 y - 30 = 0$ is tgt. to the circle. So, by

Geometry , we know that the **$\bot$ dist., say $d$, from the

Centre $O$ to the tgt. line $l$ must equal radius $r$.**

$\therefore d = r \Rightarrow | 3 \left(0\right) - 4 \left(0\right) - 30 \frac{|}{\sqrt{{\left(3\right)}^{2} + {\left(4\right)}^{2}}} = r \Rightarrow r = \frac{30}{5} = 6$

Now, the eqn. of a circle having centre at $\left(h , k\right)$ & radius $r$, is

given by, ${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$. Hence, in our case, the reqd.

eqn. is ${x}^{2} + {y}^{2} = 36$.

Note : The $\bot$dist. $d$ from a pt. $\left({x}_{1} , {y}_{1}\right)$ to a line $l : a x + b y + c = 0$

is $d = | a {x}_{1} + b {y}_{1} + c \frac{|}{\sqrt{{a}^{2} + {b}^{2}}} .$