Equation of Circle Through Origin

[mathdad]

Find the equation of the circle passing through the origin and centered at the point (3,5).

Origin means the point (0,0).
From the previous example, I found the equation of the circle centered at (3,5) to be (x - 3)^2 + (y - 5)^2 = 25.

I do not understand what part the origin plays here. The textbook does not give an example for this question. Must I find the distance from the origin to the point (3,5)?

Must I find the slope? Can I get a hint?

 

[MarkFL]

Find the equation of the circle passing through the origin and centered at the point (3,5).

Origin means the point (0,0).
From the previous example, I found the equation of the circle centered at (3,5) to be (x - 3)^2 + (y - 5)^2 = 25.

I do not understand what part the origin plays here. The textbook does not give an example for this question. Must I find the distance from the origin to the point (3,5)?

Must I find the slope? Can I get a hint?

Since the circle passes through the origin, then the distance from the origin to the center must be the radius. Now you have the radius, and the center was given, so stating the circle's equation will follow from that. :)

 

[mathdad]

d = sqrt{(3-0)^2 + (5-0)^2}

d = sqrt{3^2 + 5^2}

d = sqrt{9 + 25}

d = sqrt{34} = radius = r

The equation must be (x - 3)^2 + (y - 5)^2 = [sqrt{34}]^2, which becomes (x - 3)^2 + (y - 5)^2 = 34.

Correct?

 

[MarkFL]

One definition for a circle is the locus of all points $(x,y)$ whose distance from some central point $(h,k)$ is the same, which is called the radius, and we'll label this radius $r$. Thus, the distance formula gives us:

\(\displaystyle r=\sqrt{(x-h)^2+(y-k)^2}\)

And upon squaring, we obtain the familiar equation for a circle in standard form:

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

So, if we are given the center $(h,k)$ and one point $\left(x_1,y_1\right)$ said to be on the circle, then by definition, we know the distance between the center and the given point on the circle must be $r$.

 

[MarkFL]

d = sqrt{(3-0)^2 + (5-0)^2}

d = sqrt{3^2 + 5^2}

d = sqrt{9 + 25}

d = sqrt{34} = radius = r

The equation must be (x - 3)^2 + (y - 5)^2 = [sqrt{34}]^2, which becomes (x - 3)^2 + (y - 5)^2 = 34.

Correct?

Yes, that's correct. :)

 

[mathdad]

One definition for a circle is the locus of all points $(x,y)$ whose distance from some central point $(h,k)$ is the same, which is called the radius, and we'll label this radius $r$. Thus, the distance formula gives us:

\(\displaystyle r=\sqrt{(x-h)^2+(y-k)^2}\)

And upon squaring, we obtain the familiar equation for a circle in standard form:

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

So, if we are given the center $(h,k)$ and one point $\left(x_1,y_1\right)$ said to be on the circle, then by definition, we know the distance between the center and the given point on the circle must be $r$.

This is very interesting. Thanks for the information...