# How do you find the equations of the two tangents to the circle x^2 + y^2 - 2x - 6y + 6 = 0 which pass through the point P(-1,2)?

Jul 18, 2018

$3 x + 4 y - 5 = 0 \setminus \setminus$ & $\setminus \setminus x + 1 = 0$

#### Explanation:

Given equation of circle: ${x}^{2} + {y}^{2} - 2 x - 6 y + 6 = 0$ can re-written as

${\left(x - 1\right)}^{2} + {\left(y - 3\right)}^{2} = 4$

The above circle has center at $\left(1 , 3\right)$ & radius $2$

Let tangent passing through the point $\left(- 1 , 2\right)$ be drawn at the point $\left(h , k\right)$ on the circle: ${x}^{2} + {y}^{2} - 2 x - 6 y + 6 = 0$ then the point $\left(h , k\right)$ will satisfy the equation of circle as follows

${h}^{2} + {k}^{2} - 2 h - 6 k + 6 = 0 \setminus \ldots \ldots \ldots . . \left(1\right)$

Now, the line joining points $\left(h , k\right)$ & center $\left(1 , 3\right)$ will be perpendicular to the line joining the points $\left(h , k\right)$ & $\left(- 1 , 2\right)$ hence by condition of perpendicular lines we have

$\setminus \frac{k - 3}{h - 1} \setminus \times \frac{k - 2}{h - \left(- 1\right)} = - 1$

${h}^{2} + {k}^{2} - 5 k + 5 = 0 \setminus \ldots \ldots \ldots \ldots . \left(2\right)$

Now, subtracting (2) from (1) we get

${h}^{2} + {k}^{2} - 2 h - 6 k + 6 - \left({h}^{2} + {k}^{2} - 5 k + 5\right) = 0 - 0$

$k = 1 - 2 h \setminus \ldots \ldots \ldots . . \left(3\right)$

Substituting $k = 1 - 2 h$ in (2), we get

${h}^{2} + {\left(1 - 2 h\right)}^{2} - 5 \left(1 - 2 h\right) + 5 = 0$

$5 {h}^{2} + 6 h + 1 = 0$

$\left(5 h + 1\right) \left(h + 1\right) = 0$

$h = - \frac{1}{5} , - 1$

Substituting the values of $h$ in (3), we get corresponding values of $k$ as follows

$k = 1 - 2 \left(- \frac{1}{5}\right) , k = 1 - 2 \left(- 1\right)$

$k = \frac{7}{5} , 3$

Hence, the points at which tangents are drawn are $\left(- \frac{1}{5} , \frac{7}{5}\right)$ & $\left(- 1 , 3\right)$ Thus there are two tangents drawn from external point $\left(- 1 , 2\right)$

Now, the equation of tangent line joining $\left(- 1 , 2\right)$ & $\left(- \frac{1}{5} , \frac{7}{5}\right)$ is given as

$y - 2 = \setminus \frac{2 - \frac{7}{5}}{- 1 - \left(- \frac{1}{5}\right)} \left(x - \left(- 1\right)\right)$

$3 x + 4 y - 5 = 0$

Similarly, the equation of tangent line joining $\left(- 1 , 2\right)$ & $\left(- 1 , 3\right)$ is given as

$y - 2 = \setminus \frac{2 - 3}{- 1 - \left(- 1\right)} \left(x - \left(- 1\right)\right)$

$x + 1 = 0$

hence, the equations of tangent lines drawn from the external point to the given circle are

$3 x + 4 y - 5 = 0$ & $x + 1 = 0$

Jul 18, 2018

$x + 1 = 0 , \mathmr{and} , 3 x + 4 y - 5 = 0$.

#### Explanation:

Let us solve the Problem using Geometry.

For this, we suppose that the point of contact of the required

tangent through $P \left(- 1 , 2\right)$ is $Q \left(h , k\right)$ on the circle

$S : {x}^{2} + {y}^{2} - 2 x - 6 y + 6 = 0$.

$S : {\left(x - 1\right)}^{2} + {\left(y - 3\right)}^{2} = {2}^{2}$, we see that the centre is $C \left(1 , 3\right) .$

From Geometry, we know that, $C Q \bot P Q$.

Hence, "(the slope of CQ)"xx("the slope of "PQ)=-1.

$\therefore \left\{\frac{k - 3}{h - 1}\right\} \times \left\{\frac{k - 2}{h + 1}\right\} = - 1$.

$\therefore \left({k}^{2} - 5 k + 6\right) + \left({h}^{2} - 1\right) = 0 ,$

$i . e . , {h}^{2} + {k}^{2} - 5 k + 5 = 0. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left({\star}^{1}\right)$.

Also, $Q \in S . \therefore {h}^{2} + {k}^{2} - 2 h - 6 k + 6 = 0. \ldots \ldots \ldots . \left({\star}^{2}\right)$.

$\therefore \left({\star}^{1}\right) - \left({\star}^{2}\right) \Rightarrow 2 h + k - 1 = 0 , \mathmr{and} , k = 1 - 2 h \ldots \left({\star}^{3}\right)$.

Then, by $\left({\star}^{1}\right) , {h}^{2} + {\left(1 - 2 h\right)}^{2} - 5 \left(- 2 h\right) = 0$.

$\therefore 5 {h}^{2} + 6 h + 1 = 0 \Rightarrow h = - 1 , \mathmr{and} , h = - \frac{1}{5}$.

$\therefore k = 1 - 2 h = 3 , \mathmr{and} , k = \frac{7}{5}$.

Thus, there are two tangents through $P \left(- 1 , 2\right)$ that touch $S$

at ${Q}_{1} \left(- 1 , 3\right) \mathmr{and} {Q}_{2} \left(- \frac{1}{5} , \frac{7}{5}\right)$.

Their eqns. are, $P {Q}_{1} : x = - 1 , \mathmr{and} ,$

$P {Q}_{2} : \frac{y - 2}{\frac{7}{5} - 2} = \frac{x + 1}{- \frac{1}{5} + 1} , i . e . , 3 x + 4 y - 5 = 0$.