Question

sketch the graph of the following function: $x^{2}+y^{2}=16$. bonus question 5 points the equation of a circle whose endpoints of the diameter are p(8,-7) and q(4,5) hint calculate formula..and dont forget $(x - h)^{2}+(y - k)^{2}$

Explanation:

Step1: Recall circle - equation form

The standard form of the equation of a circle is $(x - h)^{2}+(y - k)^{2}=r^{2}$, where $(h,k)$ is the center of the circle and $r$ is the radius. For the equation $x^{2}+y^{2}=16$, we can rewrite it as $(x - 0)^{2}+(y - 0)^{2}=4^{2}$.

Step2: Determine the center

Comparing with the standard - form, when $h = 0$ and $k = 0$, the center of the circle is $(0,0)$.

Step3: Determine the radius

Since $r^{2}=16$, taking the square root of both sides (and considering the non - negative value for the radius), we get $r = 4$.

For the bonus question:

Step1: Find the center of the circle

The center of the circle with endpoints of the diameter $P(x_1,y_1)=(8, - 7)$ and $Q(x_2,y_2)=(4,5)$ is the mid - point of the diameter. The mid - point formula is $(\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})$. So, $h=\frac{8 + 4}{2}=6$ and $k=\frac{-7 + 5}{2}=-1$.

Step2: Find the radius

The radius $r$ is the distance from the center $(h,k)=(6,-1)$ to either $P$ or $Q$. Using the distance formula $d=\sqrt{(x_2 - x_1)^{2}+(y_2 - y_1)^{2}}$, with $(x_1,y_1)=(6,-1)$ and $(x_2,y_2)=(8,-7)$, we have $r=\sqrt{(8 - 6)^{2}+(-7+1)^{2}}=\sqrt{4 + 36}=\sqrt{40}=2\sqrt{10}$.

Step3: Write the equation of the circle

The equation of the circle is $(x - 6)^{2}+(y + 1)^{2}=40$.

Answer:

For $x^{2}+y^{2}=16$:
Center: $(0,0)$
Radius: $4$
For the bonus question:
Equation of the circle: $(x - 6)^{2}+(y + 1)^{2}=40$