Question

A circle with center (-2,6) passes through the point (-2,10). Which of the following is the equation of the circle?

Answer 1

Because the center is at (-2,6) its either  a or c    ( because of the x+2 and x-6)

If we plug x = -2 and y = 10 into  equation a and c well see which one is correct
Plug in  equation a:_
((-2+2)^2 + (10-6)^2 = 0 + 16 = 16 
so its equation a Answer

Answer 2

First option is correct. The equation of the  is $(x-2)^{2} +(y-6)^{2} =16$.

What is circle?

A circle is a closed two-dimensional figure in which the set of all the points in the plane is equidistant from a given point called "center".

General equation for circle

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

where,

(h, k) is the coordinate of the center of a circle

(x, y) are the coordinate points

r is the radius of circle.

According to the question

we have

the center of the circle,  (h, k) = (-2, 6)

and the circle passes through the point (-2, 10)

since, the circle passes through point (-2, 10), the radius(r) of the circle is the distance between the points (-2, 6) and  (-2, 10)

⇒ $r =\sqrt{(-2+2)^{2 }+ (6-10)^{2} }$

⇒ r = $\sqrt{0^{2}+(-4)^{2} }$

⇒$r = \sqrt{16}$

⇒$r = 4unit$

therefore,

the equation of the circle is given by

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

$(x-(-2))^{2} +(y-6)^{2} =16$

$(x+2)^{2} +(y-6)^{2} =16$ ..... (1)

$x^{2} +4+4x+y^{2} +12y+36 = 16$

$x^{2} +y^{2} +4x+12y+24=0$

Hence, equation 1 represent the equation of circle i.e. $(x+2)^{2} + (y-6)^{2} =16$

Learn more about the equation of circle here:

https://brainly.com/question/10618691

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