Wize High School Geometry Textbook (Common Core) > Circles

Equation of a Circle

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Equation of a Circle around the Centre (Origin)
The formula of a circle on the Cartesian Plane wth centre at the origin O=(0,0)O=\left(0,0\right)O=(0,0) is

x2+y2=r2\Large\boxed{x^2+y^2=r^2}x2+y2=r2​

Wize Tip
Recall the radius (r)\left(r\right)(r) is the distance from the centre of the circle to the edge. It is the same everywhere in the circle!

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Example: Equation of a Circle

a) What is the equation of a circle about the origin O=(0,0)O=\left(0,0\right)O=(0,0) and has a radius of 4?

x2+y2=r2x^2+y^2=r^2x2+y2=r2
x2+y2=42x^2+y^2=4^2x2+y2=42

ANSWER: x2+y2=16x^2+y^2=16x2+y2=16

B) What is the radius of a circle about the origin O=(0,0)O=\left(0,0\right)O=(0,0) and the equation x2+y2=81x^2+y^2=81x2+y2=81

r2=81r2=81r=9\begin{array}{ccc}
r^2&=&81\\
\sqrt{r^2}&=&\sqrt{81}\\
r&=&9
\end{array}r2r2​r​===​8181​9​

ANSWER: The radius is 999

C) What is the radius of a circle about the origin O=(0,0)O=\left(0,0\right)O=(0,0) and  passing through (−4,3)\left(-4,3\right)(−4,3)

x2+y2=r2(−4)2+(3)2=r216+9=r225=r225=r5=r\begin{array}{ccc}
x^2+y^2&=&r^2\\
(-4)^2+(3)^2&=&r^2\\
16+9&=&r^2\\
25&=&r^2\\
\sqrt {25}&=&\sqrt{r}\\
5&=&r

\end{array}x2+y2(−4)2+(3)216+92525​5​======​r2r2r2r2r​r​

ANSWER: The radius is 555

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Example: Properties of Circles
A circle has equation x2+y2=25x^2+y^2=25x2+y2=25.

a) Show that the points A(−5,0)A(-5,0)A(−5,0) and B(3,−4)B(3,-4)B(3,−4) lie on this circle.

Substitute point A into the circle equation:
Left side of the equation: x2+y2=(−5)2+(0)2=25x^2+y^2=(-5)^2+(0)^2=25x2+y2=(−5)2+(0)2=25
Right side of the equation: 252525
Since the left side of the equation equals the right sides of the equation, the point A is on the circle

Substitute point B into the circle equation:
Left side of the equation: x2+y2=(3)2+(−4)2=25x^2+y^2=(3)^2+(-4)^2=25x2+y2=(3)2+(−4)2=25
Right side of the equation: 252525
Since the left side of the equation equals the right sides of the equation, the point B is on the circle

PAGE BREAK
b) A chord is a line segment joining two points on the circumference of a circle. Find the equation of the perpendicular bisector of the chord AB.

The perpendicular bisector of the chord AB will be perpendicular to AB (have a negative reciprocal slope), and bisect AB (passes through the midpoint of AB)

Slope of the perpendicular bisector
mAB=y2−y1x2−x1=(−4)−03−(−5)=−48=−12\begin{array}{rl}
m_{AB}=&\dfrac{y_2-y_1}{x_2-x_1}\\[1em]
=&\dfrac{(-4)-0}{3-(-5)}\\[1em]
=&\dfrac{-4}{8}\\[1em]
=&-\dfrac{1}{2}
\end{array}mAB​====​x2​−x1​y2​−y1​​3−(−5)(−4)−0​8−4​−21​​

The perpendicular slope is m⊥=2m_\perp=2m⊥​=2

Midpoint of the chord AB
MAB=(x1+x22,y1+y22)=((−5)+32,0+(−4)2)=(−1,−2)\begin{array}{rl}
M_{AB}=&\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)\\[1em]
=&\left(\dfrac{(-5)+3}{2},\dfrac{0+(-4)}{2}\right)\\[1em]
=&(-1,-2)\\[1em]
\end{array}MAB​===​(2x1​+x2​​,2y1​+y2​​)(2(−5)+3​,20+(−4)​)(−1,−2)​

So, the equation of the perpendicular bisector is y−y1=m(x−x1)y−(−2)=2(x−(−1))y+2=2x+2y=2x\begin{array}{rcl}
y-y_1&=&m(x-x_1)\\
y-(-2)&=&2(x-(-1))\\
y+2&=&2x+2\\
y&=&2x
\end{array}y−y1​y−(−2)y+2y​====​m(x−x1​)2(x−(−1))2x+22x​

PAGE BREAK
c) Show that the perpendicular bisector found in part b) passes through the origin (0,0).

Substitute the point (0,0)(0,0)(0,0) into the equation of the perpendicular bisector y=2xy=2xy=2x:
Left side of the equation: y=0y=0y=0
Right side of the equation: 2x=2(0)=02x=2(0)=02x=2(0)=0
Since the left and right sides equal, the perpendicular bisector does pass through the origin.

Practice: Equation of a Circle around the Centre (Origin)
a) What is the equation of a circle about the origin O=(0,0)O=\left(0,0\right)O=(0,0) and a radius of 666?

b) What is the radius of a circle about the origin O=(0,0)O=\left(0,0\right)O=(0,0) and the equation x2+y2=23x^2+y^2=23x2+y2=23

c) What is the radius of a circle about the origin O=(0,0)O=\left(0,0\right)O=(0,0) and  passing through (5,−3)\left(5,-3\right)(5,−3)

a) Enter the equation in the form

x^2+y^2=r^2

b) Enter the exact radius (do not round your answer)

c) Enter the exact radius (do not round your answer)

I don't know

Practice: Area of a Circle
A race track is in the shape of a perfect circle. The outer edge of the track has equation x2+y2=49x^2+y^2=49x2+y2=49 and the inner edge of the track has equation x2+y2=9x^2+y^2=9x2+y2=9. Determine the area that this track covers.

Enter the exact area the track covers (do not round your answer), do not include units.

I don't know

Practice: Properties of Circles
A circle has equation x2+y2=25x^2+y^2=25x2+y2=25. The points P(0,5), Q(−5,0), R(−3,4), S(−4,−3)P(0,5),~Q(-5,0),~R(-3,4),~S(-4,-3)P(0,5), Q(−5,0), R(−3,4), S(−4,−3) all lie on this circle. Find the point of intersection of the chords PQ and RS.

Enter the exact point of intersection in the form

(x,y)

(do NOT round your answer)

I don't know