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# Find equation of a circle that passes three vertices of a triangle

Question:

The vertices of a triangle ABC have coordinates A (3, 6), B (1, 2), and C (3, 2). Find equation of the circle that passes through these three points.

Solution:

The point B and point C have the same y-coordinate. The point A and the point C have the same x-coordinate. So, the angle C is 90o. So, the triangle ABC is a right triangle. By the property of a right triangle, in a right triangle, the median on the hypotenuse is equal to half of the hypotenuse.

Let M is the midpoint of AB. M has coordinate (xo, yo). Connect MC, then MC is median on the hypotenuse. Then MC = (1/2) AB = MB = MA. So, M (xo, yo) is the center of the circle. The radius of the circle is r = MB = MA = MC. Now we will find the center of the circle M (xo, yo) and the radius of the circle r.

By the midpoint formula, xo = (1/2) (x1 + x2). yo = (1/2) (y1 + y2). Let point B is the first point and point A is the second point. Then x1 = 1 and y1 = 2. x2 = 3 and y2 = 6.

xo = (1/2) (x1 + x2)
= (1/2) (1 + 3)
= 4/2
= 2
yo = (1/2) (y1 + y2)
= (1/2) (2 + 6)
= 8/2
= 4
So, M has the coordinate (2, 4)

Now we find the radius of the circle. By the distance formula, r = BM. Let B be the first point and M be the second point. Then r = BM = square root of (x2 - x1)2 + (y2 - y1)2.

r = BM = square root of (x2 - x1)2 + (y2 - y1)2
= square root of (xo - x1)2 + (yo - y1)2
= square root of (2 - 1)2 + (4 - 2)2
= square root of (12 + 22)
= square root of (1 + 4)
= square root of 5

So, the circle equation is: (x - 2)2 + (y - 4)2 = 5

Look the figure above, the blue curve is the graph of the circle (x - 2)2 + (y - 4)2 = 5. The circle passes the vertices of the triangle ABC. Point M is the center of the circle. M has coordinate (2, 4) and the radius of the circle is square root of 5. Watch the video for more details.