
Find the equation of a circle with center (3,4) and point P (-1,3)
Click input boxes to enter data
Center (3,4)
( h ,
k )
Point (-1,3) on the radius
( x ,
y )
ANSWER is in generalized equation form of a circle which is: (x - h ) 2 + (y - k) 2 = r 2
First step is to solve for the value of r 2 . How to do that ?
Simply use the given value of x, y and h, k and simplify it to get the value of r 2
(-1-3) 2 + (3-4) 2 = r 2
(-4) 2 + (-1) 2 = r 2
16 + 1 = r 2
You can stop from here as your answer to equation of the circle.
( x - ) 2 + ( y - ) 2 = r 2 This radius, is still in the form r2. So take the square root of r,
√ r
to find the actual radius. Click the square root button
OF
=
radius
But if the requested solution is equation of a circle in simplified form then you need to expand the two binomial (x - h ) 2 and (y - k ) 2 then simplified the equation.
ANSWER after doing the binomial expansion of the above equation.
x2 - x + y2 - y + + =
ANSWER after simplifying the above equation
x2 - x + y2 - y + =
Another method to find the equation of a circle is by using Geogebra software
| Step 1. Delete point A, B, and C by selecting the three vertical dots. Then select delete |
| Step 2. Enter in input box as is (3,4) next press enter tab |
| Step 3. Enter in input box as is (-1, 3) next press enter tab. Move your graph to the left so you can see your two points. By clicking anywhere in your graph and continue pressing while dragging your mouse pointer to the left. |
| Step 4. Select circle icon. Next select circle with center through point. |
| Step 5. In your graph select point A as your center point. Next click point B. Equation of a circle will be displayed automatically on the left side. |
| You can also find the equation of any circle if you know its center point and the given radius. How ? Using Geogebra software, select circle with center and radius. That is your challenge question to learn how to do it given your acquired knowledge from this tutorial.
Another challenge question is to find the equation of any circle given the three points on the circle's circumference. |
created by Apolinario "Sam" Ortega using Geogebra 3/24/2018