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

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

(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.

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.

To find the coordinates of the center of the circle, we first need to find the midpoint of the line segment PQ. The midpoint of a line segment is calculated by taking the average of the x-coordinates and the average of the y-coordinates of the two endpoints.

Coordinates of the midpoint M = ((x1 + x2) / 2, (y1 + y2) / 2)

Coordinates of P = (2, 1)
Coordinates of Q = (4, -5)

Using the above formula, we can find the coordinates of M.
Coordinates of M = ((2 + 4) / 2, (1 + (-5)) / 2)
= (6 / 2, -4 / 2)
= (3, -2)

So, the center of the circle is C(3, -2).

To find the equation of the circle, we need to know the radius. Since line 2x - y - 13 = 0 is a tangent to the circle at point Q(4, -5), we can use the formula for the distance from a point to a line.

The distance formula from a point (x1, y1) to a line Ax + By + C = 0 is given by
Distance = |Ax1 + By1 + C| / sqrt(A^2 + B^2)

For line 2x - y - 13 = 0, the values of A, B, and C are 2, -1, and -13 respectively.

The distance from the center C(3, -2) to the line 2x - y - 13 = 0 is the radius of the circle.

Radius = |2(3) - (-2) - 13| / sqrt(2^2 + (-1)^2)
= |6 + 2 - 13| / sqrt(4 + 1)
= |8 - 13| / sqrt(5)
= 5 / sqrt(5)
= sqrt(5)

The equation of a circle with center (h, k) and radius r is given by
(x - h)^2 + (y - k)^2 = r^2

Substituting the values, we have
(x - 3)^2 + (y + 2)^2 = (sqrt(5))^2
(x - 3)^2 + (y + 2)^2 = 5

So, the equation of the circle is (x - 3)^2 + (y + 2)^2 = 5.

To sketch the circle, plot the center C(3, -2) and mark a point on the circle using the radius sqrt(5), in this case, (3, -2 + sqrt(5)).
Then draw a smooth curve passing through the center and the marked point, ensuring that the distance between any point on the curve and the center is always equal to sqrt(5).