What is the standard equation of the circle that passes through the points (-2,4),(3,-1),(6,8)?

justaguide | College Teacher | (Level 2) Distinguished Educator

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The equation of a circle is (x - a)^2 + (y - b)^2 = r^2

As we know three points through which the circle passes, we can create three equations.

(-2 , 4)

(-2 - a)^2 + (4 - b)^2 = r^2 ...(1)

(3 , -1)

(3 - a)^2 + (-1 - b)^2 = r^2 ...(2)

(6 , 8)

(6 - a)^2 + (8 - b)^2 = r^2 ...(3)

(1) - (2)

=> (-2 - a)^2 + (4 - b)^2 - (3 - a)^2 - (-1 - b)^2 = 0

=> (-2 - a - 3 + a)(-2 - a + 3 - a) + (4 - b - 1 - b)(4 - b + 1 +b) = 0

=> -5(1 - 2a) + (3 - 2b)(5) = 0

=> 2a - 1 + 3 - 2b = 0

=> a - b + 1 = 0

(2) - (3)

=> (3 - a)^2 + (-1 - b)^2 - (6 - a)^2 - (8 - b)^2 = 0

=> (3 - a + 6 - a)(3 - a - 6+ a) + ( - 1 - b - 8 + b)(-1 - b + 8 - b) =0

=> (9 - 2a)(-3) + -9( 7 - 2b) = 0

=> 9 - 2a + 21 - 6b = 0

=> 2a + 6b - 30 = 0

=> a + 3b - 15 = 0

Using a - b + 1 = 0

=> a = b - 1

b - 1 + 3b - 15 = 0

=> -4b - 16 = 0

=> b = 4

a = 3

(-2 - a)^2 + (4 - b)^2 = r^2

=> (-2 - 3)^2 + ( 4 - 4)^2 = r^2

=> 5^2 = r^2

=> r = 5

The equation of the circle is (x - 3)^2 + (y - 4)^2 = 5^2

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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The standard equation of the circle is:

x^2 + y^2 + Ax + By + C = 0

Since the circle is passing through the given points, we'll create the equations:

For the point  (-2,4):

-2A + 4B + C + 4 + 16 = 0

-2A + 4B + C + 20 = 0

2A - 4B - C = 20 (1)

For the point  (3,-1):

3A - B + C + 9+1 = 0

3A - B + C = -10 (2)

For the point  (6,8):

6A + 8B + C = -100 (3)

We'll add (1) and (2) to eliminate C:

5A - 5B = 10 We'll divide by 5;

A - B = 2 (4)

We'll add (3) and (2) to eliminate C:

8A + 4B = -80 We'll divide by 4:

2A + B = -20 (5)

We'll add (4) and (5) to eliminate B:

3A = -18

A = -6

We'll substitute A in  (4):

A - B = 2 => -6 - B = 2 => B = -8

We'll substitute B and A in  (2):

-18 + 8 + C = -10

C = 0

The standard equation of the given circle is: x^2 + y^2 - 6x - 8y = 0.