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

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:

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;

Simplify:

Question 7

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

We substitute into the formula:

To obtain:

We simplify to get;

Question 8

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

We substitute

To obtain;

We simplify to get:

Question 9

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

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

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

We simplify to obtain;

Simplify to get:

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:

We substitute the points to get:

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

This gives us

Simplify:

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 ;

We find the radius using the distance formula:

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

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

Simplify;

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;

to obtain:

The required equation is

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you: : 0

bye