3 years ago

Writing an Equation in Slope-Intercept Form

Mathematics

1 answer:

STALIN [3.7K]3 years ago

3 0

Answer:y=Mx+b

Step-by-step explanation:

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3 years ago

Prove or disprove that the point (√5, 12) is on the circle centered at the origin and containing the point (-13, 0). Show your w

pav-90 [236]

Using the equation of the circle, it is found that since it reaches an identity, the point (√5, 12) is on the circle.

<h3>What is the equation of a circle?</h3>

The equation of a circle of center $(x_0, y_0)$ and radius r is given by:

$(x - x_0)^2 + (y - y_0)^2 = r^2$

In this problem, the circle is centered at the origin, hence $(x_0, y_0) = (0,0)$ .

The circle contains the point (-13,0), hence the radius is found as follows:

$x^2 + y^2 = r^2$

$(-13)^2 + 0^2 = t^2$

$r^2 = 169$

Hence the equation is:

$x^2 + y^2 = 169$

Then, we test if point (√5, 12) is on the circle:

$x^2 + y^2 = 169$

$(\sqrt{5})^2 + 12^2 = 169$

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 brainly.com/question/24307696

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