Question

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Answer 1

Question1. We want to find the equation of the circle with center at (-3, 1) and through the point (2, 13).

Use the distance formula to find the radius.

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

$r=\sqrt{(2- - 3)^2+(13-1)^2}$

$r=\sqrt{5^2+12^2}$

$r=\sqrt{25+144}$

$r=\sqrt{169}$

$r = 13$

The equation of the circle is given by:

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

Where (a,b)=(-3,1) is the center and r=13 is the radius.

We substitute to get:

${(x + 3)}^{2} + ( {y - 1)}^{2} = {13}^{2}$

${(x + 3)}^{2} + ( {y - 1)}^{2} = 169$

Question 2) The given circle has equation:

${x}^{2} + {(y - 6)}^{2} = 50$

We want to find the center and radius of this circle.

We need to compare to

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

Rewriting the given equation will make it easy for us;

${(x - 0)}^{2} + {(y - 6)}^{2} = ( {5 \sqrt{2} )}^{2}$

Therefore the center is (0,6) and the radius is

$5 \sqrt{2}$

Question 3. We want to find the diameter of the circle with equation:

${(x + 4)}^{2} + ( {y - 9)}^{2} = 18$

By comparing to

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

We have

${r}^{2} = 18$

$r =3 \sqrt{2}$

By the diameter is twice the radius.

$diameter = 2 \times 3 \sqrt{2} = 6 \sqrt{2}$

Question 4. We want to find the equation of the circle with center at (-3, 0) and diameter 20.

We use the formula:

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

where (a,b)=(-3,0) is the center and r=20 is the radius.

${(x - - 3)}^{2} + {(y - 0)}^{2} = {20}^{2}$

${(x + 3)}^{2} + {y }^{2} =400$

Question 5) We want to find the center and radius of

${(x - 5)}^{2} + {(y + 2)}^{2} = 16$

We compare this equation to

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

Then we can see that:

$- a = - 5 \\ a = 5$

$- b = 2 \\ b = - 2$

${r}^{2} = 16$

$r = 4$

Therefore the center is ((5,-2) and the radius is 4.