3-19-2018 CLASS NOTES

CIRCLES

THE EARTH SEEN FROM APOLLO 17

WHAT IS A CIRCLE?

IS THE EARTH SPHERICAL?HOW WOULD WE KNOW THAT?

1

CIRCLE WITH CENTER (h, k) and RADIUS r.

(x, y) is any point anywhere on circle.They are all at the same distance from center.

HOW DO YOU EXPRESS THAT MATHEMATICALLY?

Using DISTANCE FORMULA,√

(x− h)2 + (y − k)2 = r =⇒ (x− h)2 + (y− k)2 = r2.

This is called the STANDARD FORM EQUATION OF THE CIRCLE.

That means Every point on the circle satisfies this equation

So if you want to check whether a point is on the circle,you just plug in the x and y coordinates of the point into the equation.

Can you write the standard form equation of the circlewith center at (0,0) and radius

√3 ?

Answer: (x− 0)2 + (y − 0)2 = (√

3)2

Simplifying, x2 + y2 = 3.

Is the point (1,2) on this circle?

Answer: No. 12 + 22 = 5, not 3.

2

GENERAL FORM EQUATION OF CIRCLE

Remember that the standard form equation is

(x− h)2 + (y − k)2 = r2.

Expanding this, we get

x2 − 2hx + h2 + y2 − 2ky + k2 = r2 =⇒ x2 − 2hx + y2 − 2ky + (h2 + k2 − r2) = 0.

In general, ANY EQUATION x2 + y2 + ax + by + c = 0is an equation of a circle in GENERAL form.

GOING FROM GENERAL FORM TO STANDARD FORM

BASICALLY COMPLETE THE SQUARE FOR TWO SQUARES

Write x2 + y2 + ax + by + c = 0 as (x2 + ax) + (y2 + by) = −c.Then add the squares of a/2 and b/2 to both sides.

Get (x2 + ax + (a/2)2) + (y2 + by + (b/2)2) = −c + (a/2)2 + (b/2)2.

This is just(x +

a

2

)2+

(y +

b

2

)2

=(a

2

)2+

(b

2

)2

− c.

EXAMPLE

Draw the graph of circle (approximate sketch is enough) representing the equationx2 + y2 + 2y = 3, marking the center, x and y intercepts as well as the length of the radius.

Complete squares as above to get it in standard form.Then find center and radius from that.

Note that to get intercepts you don’t necessarily need standard form.

3

Draw the graph of circle (approximate sketch is enough) representing the equationx2 + y2 + 2y = 3, marking the center, x and y intercepts as well as the length of the radius.

Solution: Collecting x and y terms we get x2 + (y2 + 2y) = 3.Completing the square for y2 + 2y

we get x2 + (y2 + 2y + 12) = 3 + 12 ⇒ x2 + (y + 1)2 = 22.So we have x− h = x and y − k = y + 1

and so the center is (h, k) = (0,−1) and the radius is r =√

22 = 2.

The intercepts are obtained by putting x = 0 and y = 0.We get the x intercepts by solving x2 + 02 + 0 = 3

which gives x = ±√

3.

We get y−intercepts by solving 02 + y2 + 2y = 3.This is same as y2 + 2y − 3 = 0

which can be factored to get (y + 3)(y − 1) = 0which gives y = −3 and y = 1.

center (0,-1)

Radius 2

X-INTERCEPTX-INTERCEPT

Y-INTERCEPT

Y-INTERCEPT

-3 -2 -1 1 2 3

-3

-2

-1

1

4

Given that the diameter of a circle has endpoints (-1,3) and (4,6) find the center and radiusof the circle using the midpoint and distance formulae respectively. Write the equation of

this circle in standard form and then the general form.

Soln: The center is the midpoint of the diameter. Midpoint formula gives center as(−1+4

2, 3+6

2) = (1.5, 4.5).

The radius is the distance from the center to any one of the end points. Picking (-1,3), weget r =

√(−1− 1.5)2 + (3− 4.5)2 =

√(−2.5)2 + (−1.5)2 =

√6.25 + 2.25 =

√8.5. From

this we get r2 = 8.5.The standard form of equaion of circle is (x− h)2 + (y − k)2 = r2. Plugging in the values

(h, k) = (1.5, 4.5) and r2 = 8.5 we get (x− 1.5)2 + (y − 4.5)2 = 8.5.Expanding the standard form equation we get the equation in general form as

x2 − 2(1.5)x + 1.52 + y2 − 2(4.5)y + 4.52 = 8.5⇒ x2 − 3x + y2 − 9y + 14.5 = 0.

5. Find the center and radius of the circle x2 + 8x + y2 + 6y = 0. Graph the circle markingthe center and radius.

Soln:First we collect x and y terms together.

(x2 + 8x) + (y2 + 6y) = 0.Completing the square (adding the square of half the x−coefficient and the y−coefficient to

both sides)we get(x2 + 8x + 42) + (y2 + 6y + 32) = 42 + 32 = 52.

This is same as (x + 4)2 + (y + 3)2 = 52.Comparing with (x− h)2 + (y − k)2 = r2 we get

x− h = x + 4 and y − k = y + 3. Also r2 = 52. Thus the center is (-4,-3) and radius is 5.Graph is BELOW

5

CENTER (-4,-3)

RADIUS 5

-8 -6 -4 -2

-8

-6

-4

-2

2

6