Question

Find the radius of the circle (xcosa+ysina-b)^2+(xsina-ycosa-c)^2

Answer 1

x^2sin^2(A) + 2xysin(A)cos(A) + y^2cos^2(A) = P^2 and x^2cos^2(A) - 2xysin(A)cos(A) + y^2sin^2(A) = q^2 

Add the left-halves together and the right halves together. Note the 2xysin(A)cos(A) terms CANCEL: 

x^2sin^2(A) + y^2cos^2(A) + x^2cos^2(A) + y^2sin^2(A) = P^2 + q^2 

Re-arrange the left side so we can group terms: 

x^2sin^2(A) + x^2cos^2(A) + y^2cos^2(A) + y^2sin^2(A) = P^2 + q^2 

Factor out the x^2 and y^2 from their respective terms: 

x^2 * (sin^2(A) + cos^2(A)) + y^2 * (cos^2(A) + sin^2(A)) = P^2 + q^2 

Since sin^2(A) + cos^2(A) = 1 by the Pythagorean Theorem, we can simplify to the solution: 

x^2 + y^2 = P^2 + q^2