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Type the correct answer in each box.
A circle is centered at the point (-7, -1) and passes through the point (8, 7).
The radius of the circle is
units. The point (-15,
) lies on this circle.

Type the correct answer in each box A circle is centered at the point 7 1 and passes through the point 8 7 The radius of the circle is units The point 15 lies o class=

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Answer:

Part 1) The radius is [tex]r=17\ units[/tex]

Part 2) The points (-15,14) and (-15,-16) lies on the circle

Step-by-step explanation:

step 1

Find the radius of the circle

we know that

The distance from the center to any point on the circumference is equal to the radius.

we have

[tex](-7,-1),(8,7)[/tex]

Find the distance

the formula to calculate the distance between two points is equal to

[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]

substitute

[tex]r=\sqrt{(7+1)^{2}+(8+7)^{2}}[/tex]

[tex]r=\sqrt{(8)^{2}+(15)^{2}}[/tex]

[tex]r=\sqrt{289}[/tex]

[tex]r=17\ units[/tex]

step 2

The point (-15,y) lies on the circle

we know that

If a ordered pair lie on the circle, them the ordered pair must satisfy the equation of the circle

The equation of the circle is

[tex](x-h)^{2}+(y-k)^{2}=r^{2}[/tex]

The center is (-7,-1) an the radius is 17 units

substitute

[tex](x+7)^{2}+(y+1)^{2}=17^{2}[/tex]

[tex](x+7)^{2}+(y+1)^{2}=289[/tex]

substitute the x-coordinate of the point and solve for the y-coordinate

For x=-15

[tex](-15+7)^{2}+(y+1)^{2}=289[/tex]

[tex](-8)^{2}+(y+1)^{2}=289[/tex]

[tex]64+(y+1)^{2}=289[/tex]

[tex](y+1)^{2}=289-64[/tex]

[tex](y+1)^{2}=225[/tex]

square root both sides

[tex](y+1)=(+/-)15[/tex]

[tex]y=-1(+/-)15[/tex]

[tex]y=-1(+)15=14[/tex]

[tex]y=-1(-)15=-16[/tex]

therefore

The points (-15,14) and (-15,-16) lies on the circle

see the attached figure to better understand the problem

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