Question

consider the circle centered at the origin and passing through the point (-5, 0). (a) give the equation of the circle. (b) for each point, decide whether or not it is on the circle. (x, y) is the point on the circle? yes no (-3, √21) (7, 0) (0, -5) (√7, 1)

Explanation:

Step1: Find the radius

The circle is centered at the origin $(0,0)$ and passes through $(-5,0)$. The radius $r$ is the distance from the origin to the point $(-5,0)$. Using the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$, here $x_1 = 0,y_1 = 0,x_2=-5,y_2 = 0$, so $r=\sqrt{(-5 - 0)^2+(0 - 0)^2}=5$.

Step2: Write the equation of the circle

The standard - form equation of a circle centered at the origin $(h,k)=(0,0)$ is $(x - h)^2+(y - k)^2=r^2$. Substituting $h = 0,k = 0,r = 5$ into the equation, we get $x^{2}+y^{2}=25$.

Step3: Check if points are on the circle

For the point $(-3,\sqrt{21})$: Substitute $x=-3$ and $y = \sqrt{21}$ into the equation $x^{2}+y^{2}$. We have $(-3)^{2}+(\sqrt{21})^{2}=9 + 21=30
eq25$, so it is not on the circle.
For the point $(7,0)$: Substitute $x = 7$ and $y = 0$ into the equation $x^{2}+y^{2}$. We have $7^{2}+0^{2}=49
eq25$, so it is not on the circle.
For the point $(0,-5)$: Substitute $x = 0$ and $y=-5$ into the equation $x^{2}+y^{2}$. We have $0^{2}+(-5)^{2}=25$, so it is on the circle.
For the point $(\sqrt{7},1)$: Substitute $x=\sqrt{7}$ and $y = 1$ into the equation $x^{2}+y^{2}$. We have $(\sqrt{7})^{2}+1^{2}=7 + 1=8
eq25$, so it is not on the circle.

Answer:

(a) $x^{2}+y^{2}=25$
(b)
$(-3,\sqrt{21})$: No
$(7,0)$: No
$(0,-5)$: Yes
$(\sqrt{7},1)$: No