Trig/Math
posted by Josh .
Find all the points having an x coordinate of 2 whose distance from the point (1,5) is 5.
so far i have
5=sqrt((12)^2+(2y)^2)
5=(3)^2
how do i simplify the rest of the equation so i can find x. do i foil it out? thanks.

Geometrically, you are looking for the intersection of a vertical line (x=2) with a circle centred at (1,5) with a radius of 5.
Line L1: x=2
Circle C1: (x(1))²+(y(5))²=5²
(x+1)²+(y+5)²=25
Substitute x=2 in C1:
(2+1)²+(y+5)²=25
(y+5)²=253²=16=4²
(y+5) = ±4
So there are two solutions, y=9 or y=1 
thank you!!

You're welcome.
Do give a check on the distance between the centre and the intersection points to complete the question.
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