What is the Center and Radius of a Circle?

[mathdad]

A. Determine the center and radius of circle.

B. Also, find the y-coordinates of the points (if any) where the circle intersects the y-axis.

View attachment 7462

 

[Joppy]

If the circle were to intersect the y-axis, then what would x be?

 

[mathdad]

If the circle were to intersect the y-axis, then what would x be?

If the circle intersects the y-axis, the value of x is 0. True?

 

[Joppy]

If the circle intersects the y-axis, the value of x is 0. True?

No. If the circle intersects the y-axis, the value of x is 0 at the point(s) of intersection.

We must be absolutely clear.

 

[mathdad]

Re: Center & Radius of Circle

No. If the circle intersects the y-axis, the value of x is 0 at the point(s) of intersection.

We must be absolutely clear.

How is part B found?

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Is my work for part A correct?

 

[HallsofIvy]

Yes, part A is correct.

Part B asks you to find the coordinates of the points where the circle intersects the y-axis. Joppy led you to the conclusion that those points must have x= 0. Now put x= 0 in the equation of the circle to determine what y is.

 

[mathdad]

Part B

Let x = 0

x^2 + (y + 1)^2 = 20

(0)^2 + (y + 1)^2 = 20

(y + 1)^2 = 20

sqrt{(y + 1)^2} = sqrt{20}

y + 1 = 2•sqrt{5}

y = 2•sqrt{5} - 1

The y-coordinate is 2•sqrt{5} - 1.

Yes?

Is one the points of intersection for the circle
(0, 2•sqrt{5}-1)?

 

[MarkFL]

At the point:

\(\displaystyle (y+1)^2=20\)

Your next step should be:

\(\displaystyle y+1=\pm\sqrt{20}=\pm2\sqrt{5}\)

Hence:

\(\displaystyle y=-1\pm2\sqrt{5}\)

And so the points of intersection of the given circle and the $y$-axis are:

\(\displaystyle \left(0,-1+2\sqrt{5}\right),\,\left(0,-1-2\sqrt{5}\right)\)

View attachment 7485

 

[MarkFL]

We should be able to convince ourselves that given the circle:

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

We then know these points are on the circle:

\(\displaystyle (h+r,k),\,(h-r,k),\,(h,k+r),\,(h,k-r)\)

 

[mathdad]

We should be able to convince ourselves that given the circle:

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

We then know these points are on the circle:

\(\displaystyle (h+r,k),\,(h-r,k),\,(h,k+r),\,(h,k-r)\)

Cool notes. Check your PM.