Point P(2,1) and Q(4,-5) lie on a circle. If

Question

line 2x-y-13=0 is a tangent to the circle at Q,
what is
(*) Coordinates of the center of the circle;
(*) Equation of the circle.
(*) Sketch out your circle.

Step

Answer 1

(a) the perpendicular bisector of the chord PQ contains the center of the circle. Since PQ has slope -3, the perpendicular will have slope 1/3. So, the line is

y-1 = 1/3 (x-2)

Similarly, the radius through Q is perpendicular to the line 2x-y=13. Its equation is

y+5 = -1/2 (x-4)

Those two lines intersect at the center of the circle.

Now just crank out the details.

Answer 2

I forgot that the perpendicular bisector passes through the midpoint of PQ: (3,-2). So, its equation is

y+2 = 1/3 (x-3)

So, the center of the circle is at (0,-3).

Now find the radius, and you're about done.