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.
[tex]r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
[tex]r=\sqrt{(2- - 3)^2+(13-1)^2}[/tex]
[tex]r=\sqrt{5^2+12^2}[/tex]
[tex]r=\sqrt{25+144}[/tex]
[tex]r=\sqrt{169}[/tex]
[tex]r = 13[/tex]
The equation of the circle is given by:
[tex] {(x - a)}^{2} + {(y - b)}^{2} = {r}^{2} [/tex]
Where (a,b)=(-3,1) is the center and r=13 is the radius.
We substitute to get:
[tex] {(x + 3)}^{2} + ( {y - 1)}^{2} = {13}^{2} [/tex]
[tex]{(x + 3)}^{2} + ( {y - 1)}^{2} = 169[/tex]
Question 2) The given circle has equation:
[tex] {x}^{2} + {(y - 6)}^{2} = 50[/tex]
We want to find the center and radius of this circle.
We need to compare to
[tex]{(x - a)}^{2} + {(y - b)}^{2} = {r}^{2} [/tex]
Rewriting the given equation will make it easy for us;
[tex]{(x - 0)}^{2} + {(y - 6)}^{2} = ( {5 \sqrt{2} )}^{2} [/tex]
Therefore the center is (0,6) and the radius is
[tex]5 \sqrt{2} [/tex]
Question 3. We want to find the diameter of the circle with equation:
[tex]{(x + 4)}^{2} + ( {y - 9)}^{2} = 18[/tex]
By comparing to
[tex]{(x - a)}^{2} + {(y - b)}^{2} = {r}^{2} [/tex]
We have
[tex] {r}^{2} = 18[/tex]
[tex]r =3 \sqrt{2} [/tex]
By the diameter is twice the radius.
[tex]diameter = 2 \times 3 \sqrt{2} = 6 \sqrt{2} [/tex]
Question 4. We want to find the equation of the circle with center at (-3, 0) and diameter 20.
We use the formula:
[tex]{(x - a)}^{2} + {(y - b)}^{2} = {r}^{2} [/tex]
where (a,b)=(-3,0) is the center and r=20 is the radius.
[tex]{(x - - 3)}^{2} + {(y - 0)}^{2} = {20}^{2} [/tex]
[tex]{(x + 3)}^{2} + {y }^{2} =400[/tex]
Question 5) We want to find the center and radius of
[tex]{(x - 5)}^{2} + {(y + 2)}^{2} = 16[/tex]
We compare this equation to
[tex]{(x - a)}^{2} + {(y - b)}^{2} = {r}^{2} [/tex]
Then we can see that:
[tex] - a = - 5 \\ a = 5[/tex]
[tex] - b = 2 \\ b = - 2[/tex]
[tex] {r}^{2} = 16[/tex]
[tex]r = 4[/tex]
Therefore the center is ((5,-2) and the radius is 4.
