# The equation of tangent to the circle x^2+y^2=r^2 is y=mx+c ,then the point of contact of the circle and the tangent is?

Sep 8, 2017

See below.

#### Explanation:

The point of contact can be found by equating the equations together. This will find a value that is common to both equations and consequently will be the point or points of contact.

So we have ${x}^{2} + {y}^{2} = {r}^{2}$

rearranging: $y = \sqrt{{r}^{2} - {x}^{2}}$

Tangent line: $y = m x + c$

Point of contact: $m x + c = \sqrt{{r}^{2} - {x}^{2}}$

Sep 9, 2017

$\left(1\right) : \text{The tgt. is, } y = m x \pm r \sqrt{{m}^{2} + 1} .$

$\left(2\right) : \text{The Point of contact is, } \left(\frac{- m c}{\sqrt{{m}^{2} + 1}} , \frac{c}{\sqrt{{m}^{2} + 1}}\right) ,$

where, c=pmsqrt(m^2+!).

#### Explanation:

Let, the Line $t : y = m x + c$ be tangent (tgt.) to the Circle

$S : {x}^{2} + {y}^{2} = {r}^{2.}$

Clearly, $O \left(0 , 0\right)$ is the Centre and $r$ is the Radius of $S$.

From Geometry, we know that, in a circle, the $\bot -$dist. from

the centre to a tgt. equals the radius.

Recall that, the $\bot -$dist. from pt.

$\left(h , k\right) \text{ to line } a x + b y + c = 0 ,$ is $| a h + b k + c \frac{|}{\sqrt{{a}^{2} + {b}^{2}}} .$

Utilising these facts, we have,

$| m \left(0\right) - 0 + c \frac{|}{\sqrt{{m}^{2} + 1}} = r , \mathmr{and} , c = \pm r \sqrt{{m}^{2} + 1} \ldots \ldots \ldots . \left(\ast\right) .$

Hence, the tgt. $t$ is, $y = m x \pm r \sqrt{{m}^{2} + 1} .$

If $P$ is the pt. of contact, from Geometry, we know that,

$O P \bot t , \mathmr{and} , O P \cap t = \left\{P\right\} .$

So, $P$ can be obtained by solvig the eqns. of $t \text{ and, line } O P .$

To find the eqn. of $O P$, we note that,

its slope is -1/m, &, O in OP.

$\therefore O P : y - 0 = - \frac{1}{m} \left(x - 0\right) , i . e . , y = - \frac{1}{m} \cdot x \ldots \ldots \ldots \ldots \left(1\right) .$

$t : y = m x + c \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left(2\right) .$

Solving $\left(1\right) , \mathmr{and} , \left(2\right) , x = \frac{- m c}{\sqrt{{m}^{2} + 1}} , y = \frac{c}{\sqrt{{m}^{2} + 1}} .$

Therefore, the pt. of contact is,$\left(\frac{- m c}{\sqrt{{m}^{2} + 1}} , \frac{c}{\sqrt{{m}^{2} + 1}}\right) ,$

$\text{where, } c = \pm r \sqrt{{m}^{2} + 1} .$

Enjoy Maths.!