MHB Finding the Equation of a Circle Given the Diameter Coordinates

[mathdad]

Determine the equation of the circle in standard form, given the coordinate of the diameter PQ.

P(1, -3) and Q(-5, -5)

View attachment 7471

 

[Greg]

You are asked to find the equation of a circle with diameter $\overline{PQ}$. To do that, find the distance between $P$ and $Q$ and divide that result by 2 - that will give you the radius of the circle. To find the center, find the midpoint of $\overline{PQ}$ - that's the $x$ and $y$ coordinates of the center. Does that make sense?

Standard form is $(x-h)^2+(y-k)^2=r^2$, where $h,k$ are the $x$ and $y$ coordinates of the circle's center and $r$ is the radius.

 

[HallsofIvy]

Seriously? You were asked to find the equation of the circle having those two points as endpoints of a diameter. Instead, you found the equation of the line though those points!

The two points you are given are (1, -3) and (-5, -5). The midpoint of that segment is ((1-(-5))/2, ((-3-(-5))/2)= (3, 1). That is the center of the circle. The radius of the circle is the distance from either (1, -3) or (-5, -5) to (3, 1) or, equivalently, half the distance from (1, -3) to (-5, -5), \sqrt{(1- (-5))^2+ (-3-(-5))^2}= \sqrt{40}= 2\sqrt{10}.

 

[mathdad]

We want an equation in the form (x - h)^2 + (y - k)^2 = r^2.

Yes?

 

[mathdad]

r = 2•sqrt{10}

Midpoint = (3,1)

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

Let h = 3

Let k = 1

(x - 3)^2 + (y - 1)^2 = [2•sqrt{10}]^2

(x- 3)^2 + (y - 1)^2 = 4•10

(x - 3)^2 + (y - 1)^2 = 40

 

[skeeter]

r = 2•sqrt{10}

Midpoint = (3,1)

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

Let h = 3

Let k = 1

(x - 3)^2 + (y - 1)^2 = [2•sqrt{10}]^2

(x- 3)^2 + (y - 1)^2 = 4•10

(x - 3)^2 + (y - 1)^2 = 40

try again ...

$(x+2)^2+(y+4)^2=10$

 

[mathdad]

try again ...

$(x+2)^2+(y+4)^2=10$

Are you saying that the midpoint is not (3, 1)?

Are you saying that r does not equal 2•sqrt{10}?

 

[skeeter]

Are you saying that the midpoint is not (3, 1)?

Are you saying that r does not equal 2•sqrt{10}?

that’s what my equation says ...

center $(-2,-4)$, radius $r = \sqrt{10}$

recommend you graph your equation and mine using the desmos app to verify ...

 

[MarkFL]

Determine the equation of the circle in standard form, given the coordinate of the diameter PQ.

P(1, -3) and Q(-5, -5)

Center of the circle is the midpoint of $\overline{PQ}$:

$$(h,k)=\left(\frac{(1)+(-5)}{2},\frac{(-3)+(-5)}{2}\right)=(-2,-4)$$

Radius of the circle is 1/2 of the diameter (the length of $\overline{PQ}$ or the distance from $P$ to $Q$):

$$r=\frac{1}{2}\sqrt{((1)-(-5))^2+((-3)-(-5))^2}=\frac{1}{2}\sqrt{6^2+2^2}=\frac{1}{2}\sqrt{40}=\sqrt{10}$$

Thus, the equation of the circle is:

$$(x-(-2))^2+(y-(-4))^2=(\sqrt{10})^2$$

or:

$$(x+2)^2+(y+4)^2=10$$

View attachment 7486

 

[mathdad]

Center of the circle is the midpoint of $\overline{PQ}$:

$$(h,k)=\left(\frac{(1)+(-5)}{2},\frac{(-3)+(-5)}{2}\right)=(-2,-4)$$

Radius of the circle is 1/2 of the diameter (the length of $\overline{PQ}$ or the distance from $P$ to $Q$):

$$r=\frac{1}{2}\sqrt{((1)-(-5))^2+((-3)-(-5))^2}=\frac{1}{2}\sqrt{6^2+2^2}=\frac{1}{2}\sqrt{40}=\sqrt{10}$$

Thus, the equation of the circle is:

$$(x-(-2))^2+(y-(-4))^2=(\sqrt{10})^2$$

or:

$$(x+2)^2+(y+4)^2=10$$

Great job as always.