Question

Find an equation of the circle that has center (6, 2) and passes through (-2,-6).

Answer 1

The equation of the circle with center (6, 2) and passing through (-2, -6) is found using the distance formula to calculate the radius; the final equation is (x - 6)^2 + (y - 2)^2 = 128.

Step-by-step explanation:

To find the equation of the circle with center (6, 2) and passing through the point (-2, -6), we first calculate the radius of the circle by finding the distance between the center and the point on the circle. Using the distance formula:

d = √[(x2 - x1)^2 + (y2 - y1)^2]

Substituting the given points, we get:

r = √[(-2 - 6)^2 + (-6 - 2)^2] = √[(8)^2 + (8)^2] = √[64 + 64] = √128 = 8√2

Now we use the standard equation of a circle with center (h, k) and radius r:

(x - h)^2 + (y - k)^2 = r^2

Plugging our center and radius:

(x - 6)^2 + (y - 2)^2 = (8√2)^2

(x - 6)^2 + (y - 2)^2 = 128

That is the equation of our circle.

Answer 2

(x - 6)² + (y - 2)² = 128

Step-by-step explanation:

Given as the circle has center ( 6 , 2)

and it passes through points (-2 , -6)

Now , standard equation of circle is (x -a)² + (y- b)² = r² , where (a , b) is center and r is the radius .

Again from given center the circle equation is

(x- 6)² + (y- 2)² = r²

Now this circle passes through points (-2 , -6)

so, (-2 -6)² + (-6 -2)² = r²

so, 8² + 8² = r²

Or, 128 = r²

i.e Radius pf this circle = 8
`√(2)`

Hence the circle equation formed is

(x- 6)² +( y-2)² = 128 Answer