Question

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`?

Answer 1

`R=(-2,10)`

Explanation:

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

The intersection between `y+7x+4=0` and `x-7y+47=0` is at
`I = (-3/2,13/2)` so the point `R` is at
`I = (Q+R)/2` so

`R = 2I-Q = 2(-3/2,13/2)-(-1,3) = (-2,10)`

Answer 2

`R=R(-2,10).`

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=-7x-4`.

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

`(x-2)^2+(-7x-4-7)^2=25`

`:. x^2-4x+4+49x^2+154x+121-25=0`

`:. 50x^2+150x+100=0, or, x^2+3x+2=0`

`:. (x+1)(x+2)=0`

`:. x=-1, or, x=-2`.

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

Since `(-1,3)" is already given, the reqd. pt. "R=R(-2,10).`