A circle has equation x^{2}-8x+y^{2}-6y=d. A line is tangent to this circle and passes through points A and B, (0,17) and (17,0) respectively. Find the radius of the circle.

Gradient of line: (0-17)/(17-0)= -1 equation of line: y-y₁=m(x-x₁) y-17=-1(x-0) y=17-x equation of circle: (x-4)²+(y-3)²-25=d (completing the square) (x-4)²+(y-3)²=d+25 Substitute equation of line into equation of circle:(x-4)²+(17-x-3)²=d+25, 2x²-36x+(187-d)=0 As the line is tangential to the circle, we want there to be one solution of x to this quadratic, and hence need discriminant to equal 0, (-36)²-(4)(2)(187-d)=0 8d=200 d=25=> (x-4)²+(y-3)²=50 and equation of circle is of form (x-a)²+(y-b)²=r²Thus, r²= 50 and radius of circle is square root 50