Question

A circle is centered at the point (5, -4) and passes through the point (-3, 2).

Answer 1

Answer:

$(x-5)^2+(y+4)^2 = 10^2$

Step-by-step explanation:

The formula for the equation of a circle centered at the origin (0, 0) is $x^2+y^2 = r^2$ but when the center of the circle is not the origin, then you move it back to the center mathematically.

So we have:

$(x-5)^2+(y+4)^2 = r^2$

Finding the radius $r$ by calculating the distance from the center to the given point:

$r= \sqrt{(5-(-3))^2+(-4-2)^2} = \sqrt{64+36} = \sqrt{100} = 10$

Therefore, the equation of this circle will be:

$(x-5)^2+(y+4)^2 = 10^2$

Answer 2

By definition, the equation of a circle, centered at point (a, b) is

$(x-a) ^ 2 + (y-b) ^ 2 = r ^ 2$

Where r is the radius of the circumference, and is calculated as the distance from the center to any point belonging to the circumference.

We have the center (5, -4) and the point through which the circumference passes (-3, 2), then:

$a = 5\\b = -4\\r = \sqrt {(- 3-5) ^ 2 + (2 + 4) ^ 2}\\r = 10$

Finally, the equation of the circumference is:

$(x-5) ^ 2 + (y + 4) ^ 2 = 10 ^ 2$