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Can someone help me with this please?
Right triangle ABC is located at A (−1, −2), B (−1, 1), and C (−5, 1) on a coordinate plane. What is the equation of a circle A with radius segment AC?
a.(x + 1)2 + (y + 2)2 = 9
b. (x + 5)2 + (y − 1)2 = 16
c. (x + 1)2 + (y + 2)2 = 25
d. (x + 5)2 + (y − 1)2 = 25



Answer :

jacob193

Answer:

[tex](x + 1)^{2} + (y + 2)^{2} = 25[/tex].

Step-by-step explanation:

If the center of a circle is [tex](x_{0},\, y_{0})[/tex], and the radius of the circle is [tex]r[/tex], the equation of the circle would be:

[tex](x - x_{0})^{2} + (y - y_{0})^{2} = r^{2}[/tex].

In this question:

  • The center of the circle is point [tex]{\sf A}[/tex]. The coordinates of this point are given: [tex]x_{0} = (-1)[/tex], [tex]y_{0} = (-2)[/tex],
  • It is given that the radius of this circle is equal to the length of the segment [tex]{\sf AC}[/tex]. While the exact length of this segment is not given, the coordinates of [tex]{\sf A}[/tex] and [tex]{\sf C}[/tex] are given.

To find the length of this segment, apply the distance formula: the distance between two points in a cartesian plane, [tex](x_{0},\, y_{0})[/tex] and [tex](x_{1},\, y_{1})[/tex], is:

[tex]\displaystyle \sqrt{(x_{1} - x_{0})^{2} + (y_{1} - y_{0})^{2}}[/tex].

Using this formula, the length of segment [tex]{\sf AC}[/tex] would be:
[tex]\begin{aligned} & \sqrt{(x_{1} - x_{0})^{2} + (y_{1} - y_{0})^{2}} \\ =\; & \sqrt{((-5) - (-1))^{2} + ((1 - (-2))^{2}} \\ =\; & \sqrt{25}\\ =\; & 5\end{aligned}[/tex],

The radius of the circle is equal to the length of this segment.

Hence, the equation for this circle would be:

[tex](x - x_{0})^{2} + (y - y_{0})^{2} = r^{2}[/tex].

[tex](x - (-1))^{2} + (y - (-2))^{2} = 5^{2}[/tex].

[tex](x + 1)^{2} + (y + 2)^{2} = 25[/tex].