Finding Equation of Circle

[bluefish2fish]

1. Find the equation of a circle tangent to both axis with its center in the 4th quadrant passing through (8,-9)



2.Equation of Circle (X-a)^2+(Y-b)^2)=r^2



3. I tried solving graphically, but am not sure whether my graph is precise

Thanks for the help

 

[HallsofIvy]

Think about the fact that the circle is "tangent to both axis ". If a circle has center at (a, b) and is tangent to the y-axis, then the line from (a, b) to (0, b) (the point of tangency) is a radius- and has length a. If a circle has center at (a, b) and is tangent to the y-axis, then its radius is a.

If a circle has center at (a,b) and is tangent to the x-axis, then the line from (a, b) to (a, 0) (the point of tangency) is a radius- and has length b. If a circle has center at (a, b) and is tangent to the x-axis, then its radius is b.

Now, what if a circle has center at (a, b) and is tangent to both axes? Then you get that the radius is r= a= b. That is, you know that the general equation of a circle, (x-a)²+ (y- b)²= r², for this case becomes (x-a)²+ (y- a)²= r² so you only need to find a.

You are told that the circle passes through (8, -9) so you know that x= 8, y= -9 satisfies that equation. Put x= 8, y= -9 and solve for a. You will get a quadratic equation for a so there may be 2 correct answers.

 

[Architeuthis Dux]

!

Hello there,

I would recommend using the standard equation of a circle (x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center of the circle and r is the radius. In this case, we know that the circle is tangent to both axes and its center is in the 4th quadrant, so we can set h and k as positive values. Additionally, we know that the circle passes through the point (8,-9), so we can plug in these values to get:

(x-h)^2 + (y-k)^2 = r^2
(8-h)^2 + (-9-k)^2 = r^2

We also know that the circle is tangent to both axes, so the distance from the center to each axis is equal to the radius. This means that h = k = r. Substituting this into the equations above, we get:

(8-r)^2 + (-9-r)^2 = r^2
64 - 16r + r^2 + 81 + 18r + r^2 = r^2
2r^2 + 2r - 145 = 0

Using the quadratic formula, we can solve for r and get two possible values: r = -14.5 or r = 5. However, since we know that the radius must be positive, we can discard the negative value and conclude that r = 5.

Now, to find the center of the circle, we can plug this value of r into one of the original equations and solve for h and k:

(8-h)^2 + (-9-k)^2 = 25
64 - 16h + h^2 + 81 + 18k + k^2 = 25
h^2 + k^2 - 16h + 18k + 120 = 0

We can complete the square to get (h-8)^2 + (k+9)^2 = 25, which means that the center of the circle is at (8,-9).

In conclusion, the equation of the circle is (x-8)^2 + (y+9)^2 = 25. I would recommend double-checking your graph to make sure it matches this equation. I hope this helps!