How do I find the equation of a circle that passes through points (3, 7) and (-1, -4) and has 2x+y=1 as its diameter?

Answer 1

The line y = -2x + 1 contains the diameter of the circle (given)

The perpendicular bisector of the chord joining (3,7) and (-1,-4) also contains a diameter of the circle. (the perpendicular bisector of a chord contains the center of the circle).

Since diameters of a circle intersect at its center, find the point of intersection of those two lines. Then determine the distance from that point to (3,7) or (-1,-4). That will be the radius of the circle.

Equation of Perpendicular Bisector

Slope of chord: m = [7 - (-4)] / [ 3 - (-1)] = 11/4

Midpoint of chord: (1, 3/2), using midpoint formula for midpoint of segment knowing its endpoints.

The equation:

y - 3/2 = -(4/11)(x -1)

Determine Center of Circle

-(4/11)(x-1) + 3/2 = -2x + 1 (both of these expressions equal y; now multiply every term by 22)

-8(x-1) + 33 = -44x + 22

-8x + 41 = -44x + 22

36x = -19

x = -19/36

y = (-2)(-19/36) + 1 = 37/18

So the center of the target circle is (-19/36, 37/18).

Determine the Radius of the Circle

This will be the distance from the center to let's say (-1, -4):

R = √[(-19/36 +1)² + (37/18 +4)²]

For the equation of the circle we want R², so drop the square root and evaluate the radicand:

R² = (17/36)² + (109/18)² = 47813/1296

Equation of the Circle

(x + 19/36)² + (y - 37/18)² = 47813/1296

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