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#?

2 Answers
Cesareo R.
Sep 17, 2016

#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)#

Ratnaker Mehta
Sep 21, 2016

#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).#

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