How do you write an equation of a circle with Center (-1, 15) Point on Circle (2, 13)?

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Answer 1 · 2016-11-24T17:53:51

$(x+1)^2 + (y-15)^2= 13$

If 'r' is the radius of the circle then its equation would be

$(x+1)^2 +(y-15)^2 =r^2$

Since point (2,13) lies on the circle, it must satisfy the above equation. Thus

$(2+1)^2+(13-15)^2 =r^2$
$9+4= 13=r^2$

Thus the equation of the circle becomes
$(x+1)^2 + (y-15)^2= 13$

Answer 2 · 2016-11-24T17:53:55

Please see the explanation.

Use the standard equation of a circle:

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

where $(x,y)$ is any point on the circle, $(h,k)$ is the center, and r is the radius.

You are not given r, therefore, you must substitute the given point and the center into the standard form, and then compute value of r.

$(2 - -1)^2 + (13 - 15)^2 = r^2$

$3^2 + (-2)^2 = r^2$

$r^2 = 9 + 4$

$r^2 = 13$

$r = sqrt(13)$

The equation of the circle is:

$(x - -1)^2 + (y - 15)^2 = (sqrt(13))^2$