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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.

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.