# The line y=-7x-4 cuts the circle at (-1,3) and one other point, R. How do you find the co-ordinates of R where the equation of the circle to be (x-2)^2 + (y-7)^2 = 25?

Sep 17, 2016

$R = \left(- 2 , 10\right)$

#### Explanation:

Calling $Q = \left(- 1 , 3\right)$ and according with its equation, the circle is centered at $O = \left(2 , 7\right)$ and has radius $r = 5$. The line orthogonal to $y + 7 x + 4 = 0$ passing by $O$ is $x - 7 y + C = 0$ and $C$ must obey $2 - 7 \times 7 + C = 0$ so $C = 47$

The intersection between $y + 7 x + 4 = 0$ and $x - 7 y + 47 = 0$ is at
$I = \left(- \frac{3}{2} , \frac{13}{2}\right)$ so the point $R$ is at
$I = \frac{Q + R}{2}$ so

$R = 2 I - Q = 2 \left(- \frac{3}{2} , \frac{13}{2}\right) - \left(- 1 , 3\right) = \left(- 2 , 10\right)$

Sep 21, 2016

$R = R \left(- 2 , 10\right) .$

#### Explanation:

The point $R$ is a pt. of intersection of the given line with the given

circle. Hence, solving their eqns, we must get $R$.

From the eqn. of line, $y = - 7 x - 4$.

Sub.ing this in the eqn. of circle, we have,

${\left(x - 2\right)}^{2} + {\left(- 7 x - 4 - 7\right)}^{2} = 25$

$\therefore {x}^{2} - 4 x + 4 + 49 {x}^{2} + 154 x + 121 - 25 = 0$

$\therefore 50 {x}^{2} + 150 x + 100 = 0 , \mathmr{and} , {x}^{2} + 3 x + 2 = 0$

$\therefore \left(x + 1\right) \left(x + 2\right) = 0$

$\therefore x = - 1 , \mathmr{and} , x = - 2$.

Correspondingly, x=-1 rArr y=-7x-4=3, &, x=-2 rArr y=10.

Since $\left(- 1 , 3\right) \text{ is already given, the reqd. pt. } R = R \left(- 2 , 10\right) .$