+130554 Another way to do this (more difficult) is to use the Secant-Tangent Theorem
Note, flflvm97, the line y = x + 3 doesn't actually intersect this circle, but the line y = x + √2 does.....
Call the y coordinate of R, "y".....using symmetry the distance between the two points where the line x = .75 intersects the circle = 2y
And using y = x+ √2, the x coordinate of the intercept of the line and the circle =
x^2 + (x + √2)^2 = 1
x^2 + x^2 + 2√2x + 2 = 1
2x^2 + 2√2x + 1 = 0
2x^2 + √8x + 1 = 0 and using the quad formula x = - √8/4 = -1/√2
And using y = x + √2, (-1/ √2)^2 + y^2 = 1 so y^2 = 1/2 and y = the positive root= 1/ √2
So the intersection of the line and the circle is ( -1/√2, 1/√2)
And when x = .75, the y coordinate of this point on the line is (.75 + √2)
And using the Secant-Tangent Theorem ,we have
( .75 + √2 + y)*(.75 + √2 - y) = (.75 + 1/ √2)^2 + (.75 + √2 - 1/ √2)^2
( .75 + √2 + y)*(.75 + √2 - y) = (.75 + 1/ √2)^2 + (.75 + 1/ √2)^2
(.75 + √2)^2 - y^2 = 2 (.75 + 1/ √2)^2
So
y^2 = (.75 + √2)^2 - 2 (.75 + 1/ √2)^2
And y = ±√((.75 + √2)^2 - 2 (.75 + 1/ √2)^2) = ±.661438 = ±√(7)/4
And choose the positive root.....and this is the y coordinate of "R".....so R = (3/4, √(7)/4)
(I told you it was more difficult !!!)
