Let the centre of a circle, passing through the point (0,0),(1,0) and touching the circle x2+y2=9, be (h,k). Then for all possible values of the coordinates of the centre (h,k), 4(h2+k2) is equal to
Answer: 9
Solution
A circle through the origin is x2+y2−2hx−2ky=0; through (1,0) gives h=21. Internal tangency to x2+y2=9 means h2+k2=3−h2+k2⇒h2+k2=23, so h2+k2=49 and 4(h2+k2)=9.