Question

Right triangle ABC is located at A (−1, 4), B (−1, 1), and C (−5, 1) on a coordinate plane. What is the equation of a circle A with radius segment AC?

(x + 1)2 + (y − 4)2 = 9
(x + 5)2 + (y − 1)2 = 25
(x + 5)2 + (y − 1)2 = 16
(x + 1)2 + (y − 4)2 = 25

Answer 1

The circle is in A, so (-1, 4) coords

(x - xo)² + (y - yo)² = r²

(x - (-1))² + (y - 4)² = r²

(x + 1)² + (y - 4)² = r²

And the radius is the length of segment AC

distance of AC is

$\sqrt{(yc-ya)^2+(xc-xa)^2}$

$\sqrt{(1-4)^2+(-5-(-1))^2}$

$\sqrt{(-3)^2+(-4)^2}$

$\sqrt{9+16}$

$\sqrt{25}$

$5$

So r = 5

(x + 1)² + (y - 4)² = 5²

(x + 1)² + (y - 4)² = 25

Answer 2

The equation of a circle A with radius segment AC is $(x+1)^{2}+(y-4)^{2} = 25$.

In this question, we must determine the Length of Segment AC to know the Radius of the Circle and finally we construct of the equation of the Circle based on a formula from Analytic Geometry.  

First, we calculate the length of segment AC by the Length Formula for Line Segment:

$r = \sqrt{(x_{C}-x_{A})^{2}+(y_{C}-y_{A})^{2}}$ (1)

Where:

$x_{A},\,y_{A}$ - Coordinates of point A.

$x_{C},\,y_{C}$ - Coordinates of point C.

If we know that $A(x,y) = (-1, 4)$ and $C(x,y) = (-5, 1)$, then the radius of the circle is:

$r = \sqrt{[-5-(-1)]^{2}+(1-4)^{2}}$

$r = 5$

From Analytic Geometry, we know that Circle can be modelled by knowing both its Radius and its Center:

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

Where $(h,k)$ are the coordinates of the Center of the Circle.

If we know that $(h,k) = (-1, 4)$ and $r = 5$, then the equation of the circle is:

$(x+1)^{2}+(y-4)^{2} = 25$

The equation of a circle A with radius segment AC is $(x+1)^{2}+(y-4)^{2} = 25$.

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