Using the equation of the circle, it is found that since it reaches an identity, the point (√5, 12) is on the circle.
What is the equation of a circle?
The equation of a circle of center [tex](x_0, y_0)[/tex] and radius r is given by:
[tex](x - x_0)^2 + (y - y_0)^2 = r^2[/tex]
In this problem, the circle is centered at the origin, hence [tex](x_0, y_0) = (0,0)[/tex].
The circle contains the point (-13,0), hence the radius is found as follows:
[tex]x^2 + y^2 = r^2[/tex]
[tex](-13)^2 + 0^2 = t^2[/tex]
[tex]r^2 = 169[/tex]
Hence the equation is:
[tex]x^2 + y^2 = 169[/tex]
Then, we test if point (√5, 12) is on the circle:
[tex]x^2 + y^2 = 169[/tex]
[tex](\sqrt{5})^2 + 12^2 = 169[/tex]
25 + 144 = 169
Which is an identity, hence point (√5, 12) is on the circle.
More can be learned about the equation of a circle at https://brainly.com/question/24307696
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