Question

show your solutions.

1. what are the center and radius of

(a) $x^{2}+y^{2}=26$?
(b) $(x + 5)^{2}+(y - 4)^{2}=12$?

1. what is the equation of a circle with diameter 8 and center at the origin?

Explanation:

Step1: Recall the standard form of a circle equation

The standard - form of a circle equation is $(x - a)^2+(y - b)^2=r^2$, where $(a,b)$ is the center of the circle and $r$ is the radius.

Step2: Solve part (a) of question 1

For the equation $x^{2}+y^{2}=26$, we can rewrite it as $(x - 0)^2+(y - 0)^2=\sqrt{26}^2$. So the center $(a,b)=(0,0)$ and the radius $r = \sqrt{26}$.

Step3: Solve part (b) of question 1

For the equation $(x + 5)^2+(y - 4)^2=12$, comparing with the standard form $(x - a)^2+(y - b)^2=r^2$, we have $a=-5$, $b = 4$ and $r=\sqrt{12}=2\sqrt{3}$. So the center is $(-5,4)$ and the radius is $2\sqrt{3}$.

Step4: Solve question 2

Given the diameter $d = 8$, then the radius $r=\frac{d}{2}=4$ and the center is at the origin $(0,0)$. Using the standard - form of the circle equation $(x - a)^2+(y - b)^2=r^2$, substituting $a = 0$, $b = 0$ and $r = 4$, we get $x^{2}+y^{2}=16$.

Answer:

1. (a) Center: $(0,0)$, Radius: $\sqrt{26}$

(b) Center: $(-5,4)$, Radius: $2\sqrt{3}$

1. $x^{2}+y^{2}=16$