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
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