# Equation of a Circle

In this topic we will learn how to create an equation of a circle with different given. Being familiar of the method will boost our speed and accuracy. Most of the problems we encounter at school are just recycled.

General equation of a circle: $x^2+y^2+Cx+Dy+E=0$  where C, D and E are real numbers

Equation of a circle in center-radius form:  $(x-h)^2+(y-k)^2=r^2$  where (h,k) is the vertex of the circle and  r is the radius of
the circle

Sample Problem 1: (4th Year 2013 MMC Elimination)

Find the center and the radius of the circle $x^2-4x+y^2+14y+47=0$

Solution:

To solve this problem we need to convert this standard form to center radius form by completing the square of x and y. $x^2-4x+y^2+14y=-47$ $(x^2-4x+4 )+(y^2+14y+49)= -47+4+49$ $(x-2)^2+(y+7)^2=6$

From the equation above we can see that h=2, k=-7 and radius of √6

The desired answer is (2,-7) and $\sqrt{6}$ units

Sample Problem 2: (4th Year 2005 MMC elimination)

Write the equation of the circle tangent to y=-3 and with the center at (-2,2) in center radius form.

Solution:

We write the equation of the circle: $(x-h)^2+(y-k)^2=r^2$  , we are given already with $(h,k)$. We just need the radius.

The radius of the circle is equal to the distance between the point (-2,2) and y=-3. Since the circle passes line y=-3 it intersects the point the point (-2,-3). It is now obvious that the radius is 5 units. $(x+2)^2+(y-2)^2=5^2\leftrightarrow (x+2)^2+(y-2)^2=25$

Sample Problem 3: (4th year 2005 MMC elimination)

Find the equation of the circle in standard form concentric with x2+y2+4x-8y-12=0 and passing through (2,-1)

Solution:

Concentric circle are circles with the same center. We can get the center of the circle from x2+y2+4x-8y-12=0. By completing the square we get h=-2, k=4. To solve for the radius of the desired circle we solve for the distance between the center (-2,4) and a point on the circumference of the circle (2,-1) using the distance formula. $r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

By substitution: $r=\sqrt{(2-(-2))^2+(-1-4)^2}$ $r=\sqrt{41}$

Solving for the equation of the circle: $(x-h)^2+(y-k)^2=r^2$ $(x+2)^2+(y-4)^2=(\sqrt{41})^2$ $x^2+y^2+4x-8y-21=0$

### Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.