# Coordinate geometry of the circle question

• Ed Aboud
In summary, the conversation discussed finding the equations of two circles that pass through a given point and touch the x-axis and y-axis. The equations for circles were derived and it was concluded that the equations have complex roots. After further discussion, the correct equations were determined to be (x-R)^2 + (y+R)^2 = R^2, (x-2)^2 + (y+2)^2 = R^2, and (x-10)^2 + (y+10)^2 = R^2. The original attempt had the incorrect assumption that f=g=r, when in fact, |f|=|g|=r. The correct signs for the equations were determined based on the location of the given point

## Homework Statement

Fairly straight forward question but I just can't see what's wrong.

A circle passes through the point (2,-4) and touches both the x-axis and the y-axis. Find the equations of the two circles which satisfy these conditions.

## Homework Equations

$$x^2 + y^2 + 2gx + 2fy + c = 0$$

with a centre point $c (-g,-f)$

$$r = \sqrt{g^2 + f^2 - c}$$

## The Attempt at a Solution

After drawing a diagram I concluded that r=g and r=f therefore g=f.

$$g = \sqrt{g^2 + f^2 - c}$$

$$g^2 = g^2 + f^2 - c$$

$$f^2 = c$$

Since the point (2,-4) is on the circle it will satisfy :

$x^2 + y^2 + 2gx + 2fy + c = 0$

$$(2)^2 + (-4)^2 + 2g(2) + 2f(-4) + c = 0$$

$$4 + 16 + 4g -8f + c = 0$$

$$20 + 4g -8f + c = 0$$

But $f^2 = c$ and $f = g$

So

$$20 + 4f - 8f + f^2 = 0$$
$$20 -4f + f^2 = 0$$
$$f^2 -4f +20 = 0$$

This quadratic only has complex roots.

Thanks for any help!

If the center is $(x_0,y_0)$, that is, the equation of the circle is $(x-x_0)^2+ (y-y_0)^2= R^2$ and it "touches" the axes- by which I assume you mean it is tangent to them, then $(x_0,0)$ and $(0,y_0)$ are the points at which the circle is tangent to the axes. Putting $x= x_0$, y= 0 into the equation, $(-y_0)^2= R^2$ and putting $y= y_0$, x= 0, $(-x_0)^2= R^2$ so we must have $x_0= y_0= R$ so we can write the equation as $(x- R)^2+ (y- R)^2= R^2$. Adding the condition that (2,-4) lies on that circle gives $(2-R)^2+ (-4-R)^2= R^2$ which you can solve for R.

Yeah so you get $$R^2 + 4R + 20 = 0$$

this has complex roots as well...

The center of the circle isn't at (R,R). That's in the wrong quadrant. You have to pick y0 to be the negative root of y0^2=R^2. Put it at (R,-R).

Ok cool thanks so it is

$$R^2 -12R +20 = 0$$

$$R=2$$ and $$R=10$$

Therefore the equations of the circles are

$$(x- R)^2+ (y+ R)^2= R^2$$

$$(x- 2)^2+ (y+ 2)^2= R^2$$

and

$$(x- 10)^2+ (y+ 10)^2= R^2$$

Just out of curiosity what was wrong with my original attempt?

Ed Aboud said:
Ok cool thanks so it is

$$R^2 -12R +20 = 0$$

$$R=2$$ and $$R=10$$

Therefore the equations of the circles are

$$(x- R)^2+ (y+ R)^2= R^2$$

$$(x- 2)^2+ (y+ 2)^2= R^2$$

and

$$(x- 10)^2+ (y+ 10)^2= R^2$$

Just out of curiosity what was wrong with my original attempt?

Exactly the same thing. You said f=g=r. But that's not true. |f|=|g|=r is what is true. You have to figure out the signs based on the location of (2,-4).

## 1. What is the equation for a circle in coordinate geometry?

In standard form, the equation for a circle with center (h,k) and radius r is (x-h)^2 + (y-k)^2 = r^2.

## 2. How do you find the center and radius of a circle from its equation?

To find the center, set the equation equal to 0 and complete the square for both the x and y terms. The resulting values for h and k will be the coordinates of the center. The radius can be found by taking the square root of the constant term on the right side of the equation.

## 3. Can you graph a circle without knowing its equation?

Yes, you can graph a circle by plotting points that satisfy the equation (x-h)^2 + (y-k)^2 = r^2. The coordinates of the center (h,k) can serve as a starting point, and then you can plot points by substituting different values for x and y to find the corresponding points on the circle.

## 4. How do you determine if a point lies on a circle or inside/outside of it?

To determine if a point (x,y) lies on a circle, simply plug in the values for x and y into the equation of the circle. If the resulting equation is true, then the point lies on the circle. If the resulting equation is less than or greater than the radius squared, then the point is inside or outside the circle, respectively.

## 5. Can two circles intersect at more than two points?

No, two circles can only intersect at a maximum of two points. This is because a circle is defined by all the points that are equidistant from its center, and therefore, any more than two points of intersection would make the circles overlap or coincide.