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Circles Not Centered at the Origin

Write and graph equations of circles with centers not at (0, 0)

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Circles Centered at (h, k)

You draw a circle that is centered at \begin{align*}(-2, -2)\end{align*}(2,2). You measure the diameter of the circle to be 18 units. Does the point \begin{align*}(4, 5)\end{align*}(4,5) lie on the circle?


When a circle is centered at the origin (as in the last concept), the equation is \begin{align*}x^2+y^2=r^2\end{align*}x2+y2=r2. If we rewrite this equation, using the center, it would look like \begin{align*}(x-0)^2+(y-0)^2=r^2\end{align*}(x0)2+(y0)2=r2. Extending this idea to any point as the center, we would have \begin{align*}(x-h)^2+(y-k)^2=r^2\end{align*}(xh)2+(yk)2=r2, where \begin{align*}(h, k)\end{align*}(h,k) is the center.

Example A

Find the center and radius of \begin{align*}(x+1)^2+(y-3)^2=16\end{align*}(x+1)2+(y3)2=16 and graph.

Solution: Using the general equation above, the center would be \begin{align*}(-1, 3)\end{align*}(1,3) and the radius is \begin{align*}\sqrt{16}\end{align*}16 or 4. To graph, plot the center and then go out 4 units up, down, to the left, and to the right.

Example B

Find the equation of the circle with center \begin{align*}(2, 4)\end{align*}(2,4) and radius 5.

Solution: Plug in the center and radius to the equation and simplify.

\begin{align*}(x-2)^2+(y-4)^2&=5^2 \\ (x-2)^2+(y-4)^2&=25\end{align*}


Example C

Find the equation of the circle with center \begin{align*}(6, -1)\end{align*}(6,1) and \begin{align*}(5, 2)\end{align*}(5,2) is on the circle.

Solution: In this example, we are not given the radius. To find the radius, we must use the distance formula, \begin{align*}d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\end{align*}d=(x2x1)2+(y2y1)2.

\begin{align*}r&=\sqrt{\left(5-6\right)^2+ \left(2- \left(-1\right)\right)^2} \\ &=\sqrt{\left(-1\right)^2+3^2} \\ &=\sqrt{1+9} \\ &=\sqrt{10}\end{align*}


Therefore, the equation of this circle is \begin{align*}(x-6)^2+(y-(-1))^2=\left(\sqrt{10}\right)^2\end{align*} or \begin{align*}(x-6)^2+(y+1)^2=10\end{align*}.

Intro Problem Revisit In this lesson, you learned the equation of a circle that is centered somewhere other than the origin is \begin{align*}(x-h)^2+(y-k)^2=r^2\end{align*}, where \begin{align*}(h, k)\end{align*} is the center.

We are given that the center is \begin{align*}(-2, -2)\end{align*}, so \begin{align*}h= -2\end{align*} and \begin{align*}k = -2\end{align*}. We are also given the diameter of the circle, but we need the radius. Recall that the radius is half the diameter, so \begin{align*}r = \frac{18}{2} = 9\end{align*}.

If we plug these values into the equation for the circle, we get:

\begin{align*}(x-h)^2+(y-k)^2=r^2\\ (x-(-2))^2 + (y - (-2))^2 = 9^2\\ (x + 2)^2 + (y + 2)^2 = 81\end{align*}

Now to find if the point \begin{align*}(4, 5)\end{align*} lies on the circle we substitute 4 for x and 5 for y and see if the equation holds true.

\begin{align*}(x + 2)^2 + (y + 2)^2 = 81\\ (4 + 2)^2 + (5 + 2)^2 \overset{?}{=} 81\\ 6^2 + 7^2 \overset{?}{=} 81\\ 85 \ne 81\end{align*}

Therefore, the point does not lie on the circle.

Guided Practice

1. Graph \begin{align*}(x+4)^2+(y+3)^2=25\end{align*} and find the center and radius.

2. Find the equation of the circle with center \begin{align*}(-8, 3)\end{align*} and \begin{align*}(2, -5)\end{align*} is on the circle.

3. The endpoints of a diameter of a circle are \begin{align*}(-3, 1)\end{align*} and \begin{align*}(9, 6)\end{align*}. Find the equation.


1. The center is \begin{align*}(-4, -3)\end{align*} and the radius is 5.

2. Use the distance formula to find the radius.

\begin{align*}r&=\sqrt{\left(2- \left(-8\right)\right)^2+ \left(-5-3\right)^2}\\ &=\sqrt{10^2+ \left(-8\right)^2} \\ &=\sqrt{100+64} \\ &=\sqrt{164}\end{align*}

The equation of this circle is \begin{align*}(x+8)^2+(y-3)^2=164\end{align*}.

3. In this problem, we are not given the center or radius. We can find the length of the diameter using the distance formula and then divide it by 2.

\begin{align*}d&=\sqrt{\left(9- \left(-3\right)\right)^2+ \left(6-1\right)^2} \\ &=\sqrt{12^2+5^2} \qquad \text{The radius is} \ 13 \div 2=\frac{13}{2}. \\ &=\sqrt{144+25} \\ &=\sqrt{169}=13\end{align*}

Now, we need to find the center. Use the midpoint formula with the endpoints.

\begin{align*}c&=\left(\frac{-3+9}{2}, \frac{1+6}{2}\right) \\ &=\left(3, \frac{7}{2}\right)\end{align*}

Therefore, the equation is \begin{align*}(x-3)^2+ \left(y- \frac{7}{2}\right)^2=\frac{169}{4}\end{align*}.

Explore More

For questions 1-4, match the equation with the graph.

  1. \begin{align*}(x-8)^2+(y+2)^2=4\end{align*}
  2. \begin{align*}x^2+(y-6)^2=9\end{align*}
  3. \begin{align*}(x+2)^2+(y-3)^2=36\end{align*}
  4. \begin{align*}(x-4)^2+(y+4)^2=25\end{align*}

Graph the following circles. Find the center and radius.

  1. \begin{align*}(x-2)^2+(y-5)^2=16\end{align*}
  2. \begin{align*}(x+4)^2+(y+3)^2=18\end{align*}
  3. \begin{align*}(x+7)^2+(y-1)^2=8\end{align*}

Find the equation of the circle, given the information below.

  1. center: \begin{align*}(-3,-3)\end{align*} radius: 7
  2. center: \begin{align*}(-7,6)\end{align*} radius: \begin{align*}\sqrt{15}\end{align*}
  3. center: \begin{align*}(8, -1)\end{align*} point on circle: \begin{align*}(0, 14)\end{align*}
  4. center: \begin{align*}(-2, -5)\end{align*} point on circle: \begin{align*}(3, 2)\end{align*}
  5. diameter endpoints: \begin{align*}(-4, 1)\end{align*} and \begin{align*}(6, 3)\end{align*}
  6. diameter endpoints: \begin{align*}(5, -8)\end{align*} and \begin{align*}(11, 2)\end{align*}
  7. Is \begin{align*}(-9, 12)\end{align*} on the circle \begin{align*}(x+5)^2+(y-6)^2=54\end{align*}? How do you know?
  8. Challenge Using the steps from #16 in the previous concept, find the equation of the tangent line to the circle with center \begin{align*}(3, -4)\end{align*} and the point of tangency is \begin{align*}(-1, 8)\end{align*}.
  9. Extension Rewrite the equation of the circle, \begin{align*}x^2+y^2+4x-8y+11=0\end{align*} in standard form by completing the square for both the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} terms. Then, find the center and radius.


Standard Form

Standard Form

The standard form of a circle is (x-h)^2+(y-k)^2=r^2, where (h, k) is the center and r is the radius.

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