QUESTION IMAGE
Question
what is the center of the circle with the equation (x - 7)^2+(y + 8)^2 = 49? a. (-7,8) b. (-7,-8) c. (7,8) d. (7,-8) what is the center of the circle with the equation (x + 5)^2+(y - 6)^2 = 36? a. (-5,-6) b. (5,-6) c. (5,6) d. (-5,6) what is the radius of the circle with the equation (x + 2)^2+(y - 3)^2 = 100? a. 4 b. 5 c. 20 d. 10
Step1: Recall circle - equation formula
The standard form of a circle's equation is $(x - h)^2+(y - k)^2=r^2$, where $(h,k)$ is the center of the circle and $r$ is the radius.
First question:
For the equation $(x - 7)^2+(y + 8)^2 = 49$, we can rewrite $(y + 8)$ as $(y-(-8))$. Comparing with the standard - form $(x - h)^2+(y - k)^2=r^2$, we have $h = 7$ and $k=-8$. So the center of the circle is $(7,-8)$.
Second question:
For the equation $(x + 5)^2+(y - 6)^2 = 36$, we can rewrite $(x + 5)$ as $(x-(-5))$. Comparing with the standard - form $(x - h)^2+(y - k)^2=r^2$, we have $h=-5$ and $k = 6$. So the center of the circle is $(-5,6)$.
Third question:
For the equation $(x + 2)^2+(y - 3)^2 = 100$, comparing with the standard - form $(x - h)^2+(y - k)^2=r^2$, where $r^2 = 100$. Taking the square root of both sides, we get $r=\sqrt{100}=10$.
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- d. $(7,-8)$
- d. $(-5,6)$
- d. $10$