Question

Write the standard form of the equation for the given information.
6. center (3,2) radius 6
7. center (-4,-7) radius 8
8. center (5,-9) radius 10
9. center (-8,0) diameter 14
10.center (4,5) and point on the circle (3,-7).
11. diameter with endpoints (6,4) and (10,-8)
12.center (4,9) and tangent to the x-axis

Answer 1

Question 6

We want to find the equation of a circle with center (3,2) and radius 6.

The standard equation of a circle is given by:

${(x - a)}^{2} + {(y - b)}^{2} = {r}^{2}$

where (a,b) is the center of the circle and r is the radius.

We substitute (a=3,b=2) and r=6 to get;

$( {x - 3)}^{2} + ( {y - 2)}^{2} = {6}^{2}$

Simplify:

${(x - 3)}^{2} + {(y - 2)}^{2} = 36$

Question 7

This time the center is (a=-4,b=-7) and radius is r= 8.

We substitute into the formula:

${(x - a)}^{2} + {(y - b)}^{2} = {r}^{2}$

To obtain:

${(x - - 4)}^{2} + {(y - - 7)}^{2} = {8}^{2}$

We simplify to get;

${(x + 4)}^{2} + {(y + 7)}^{2} = 64$

Question 8

Now the center is (a=5,b=-9) radius is r=10

We substitute

${(x - a)}^{2} + {(y - b)}^{2} = {r}^{2}$

To obtain;

${(x - 5)}^{2} + {(y - - 9)}^{2} = {10}^{2}$

We simplify to get:

${(x - 5)}^{2} + {(y + 9)}^{2} =100$

Question 9

This circle has center (-8,0) diameter 14.

To find the radius, we divide the diameter into 2.

$r = \frac{14}{2} = 7$

We substitute the center and radius into the standard equation of the circle

${(x - a)}^{2} + {(y - b)}^{2} = {r}^{2}$

We simplify to obtain;

${(x - - 8)}^{2} + {(y - 0)}^{2} = {7}^{2}$

Simplify to get:

${(x + 8)}^{2} + {y }^{2} = 49$

Question 10.

This time we were given: center (4,5) and point on the circle (3,-7).

We determine the radius using the distance formula:

$r = \sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}$

We substitute the points to get:

$r = \sqrt{(3-4)^2 +( - 7 - 5)^2}$

$r = \sqrt{( - 1)^2 +( - 12)^2}$

$r = \sqrt{1 +144} \\ r = \sqrt{145}$

We substitute the center (4,5) and radius in to the standard equation

${(x - a)}^{2} + {(y - b)}^{2} = {r}^{2}$

This gives us

${(x - 4)}^{2} + {(y - 5)}^{2} = { (\sqrt{145} })^{2}$

Simplify:

${(x - 4)}^{2} + {(y - 5)}^{2} = 145$

Question 11.

This time we have diameter with endpoints (6,4) and (10,-8).

The midpoint is the center of the circle.

By the midpoint rule, the center is ;

$( \frac{10 + 6}{2} , \frac{4 + - 8}{2} ) \\ ( \frac{1 6}{2} , \frac{ - 4}{2} ) \\ ( \frac{10 + 6}{2} , \frac{4 + - 8}{2} ) \\ ( 8 , - 2)$

We find the radius using the distance formula:

$r = \sqrt{(x_2-x_1)^2 +(y_2-y_1)^2}$

Substitute the center (8,-2) and the point (6,4).

$r = \sqrt{(6 - 8)^2 +(4- - 2)^2}$

$r = \sqrt{( - 2)^2 +(6)^2} \\ r = \sqrt{4+36} \\ r = \sqrt{40}$

We substitute the center and radius into the standard equation now to get;

${(x - 8)}^{2} + {(y - - 2)}^{2} = {( \sqrt{40} })^{2}$

Simplify;

${(x - 8)}^{2} + {(y + 2)}^{2} =40$

Question 12;

We were given center (4,9) and the condition that the tangent to this circle is the x-axis.

This means the radius is 9, the distance from the x-axis to the center.

We substitute the center (4,9) and radius r=9 into;

${(x - a)}^{2} + {(y - b)}^{2} = {r}^{2}$

to obtain:

${(x - 4)}^{2} + {(y - 9)}^{2} = {9}^{2}$

The required equation is

${(x - 4)}^{2} + {(y - 9)}^{2} = 81$