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Circles Not Centered at the Origin

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Circles Centered at (h, k)

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

Guidance

When a circle is centered at the origin (as in the last concept), the equation is x^2+y^2=r^2 . If we rewrite this equation, using the center, it would look like (x-0)^2+(y-0)^2=r^2 . Extending this idea to any point as the center, we would have (x-h)^2+(y-k)^2=r^2 , where (h, k) is the center.

Example A

Find the center and radius of (x+1)^2+(y-3)^2=16 and graph.

Solution: Using the general equation above, the center would be (-1, 3) and the radius is \sqrt{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 (2, 4) and radius 5.

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

(x-2)^2+(y-4)^2&=5^2 \\(x-2)^2+(y-4)^2&=25

Example C

Find the equation of the circle with center (6, -1) and (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, d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} .

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}

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

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

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

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

(x-h)^2+(y-k)^2=r^2\\(x-(-2))^2 + (y - (-2))^2 = 9^2\\(x + 2)^2 + (y + 2)^2 = 81

Now to find if the point (4, 5) lies on the circle we substitute 4 for x and 5 for y and see if the equation holds true.

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

Therefore, the point does not lie on the circle.

Guided Practice

1. Graph (x+4)^2+(y+3)^2=25 and find the center and radius.

2. Find the equation of the circle with center (-8, 3) and (2, -5) is on the circle.

3. The endpoints of a diameter of a circle are (-3, 1) and (9, 6) . Find the equation.

Answers

1. The center is (-4, -3) and the radius is 5.

2. Use the distance formula to find the radius.

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}

The equation of this circle is (x+8)^2+(y-3)^2=164 .

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.

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

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

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

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

Vocabulary

Standard Form (of a Circle)
(x-h)^2+(y-k)^2=r^2 , where (h, k) is the center and r is the radius.

Explore More

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

  1. (x-8)^2+(y+2)^2=4
  2. x^2+(y-6)^2=9
  3. (x+2)^2+(y-3)^2=36
  4. (x-4)^2+(y+4)^2=25

Graph the following circles. Find the center and radius.

  1. (x-2)^2+(y-5)^2=16
  2. (x+4)^2+(y+3)^2=18
  3. (x+7)^2+(y-1)^2=8

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

  1. center: (-3,-3) radius: 7
  2. center: (-7,6) radius: \sqrt{15}
  3. center: (8, -1) point on circle: (0, 14)
  4. center: (-2, -5) point on circle: (3, 2)
  5. diameter endpoints: (-4, 1) and (6, 3)
  6. diameter endpoints: (5, -8) and (11, 2)
  7. Is (-9, 12) on the circle (x+5)^2+(y-6)^2=54 ? 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 (3, -4) and the point of tangency is (-1, 8) .
  9. Extension Rewrite the equation of the circle, x^2+y^2+4x-8y+11=0 in standard form by completing the square for both the x and y terms. Then, find the center and radius.

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