QUESTION IMAGE
Question
write the standard form of the equation and the general form of the equation of the circle with radius r and center (h,k). then graph the circle. r = 2; (h,k)=(-4,-3). the standard form of the equation of this circle is . (type your answer in standard form.) the general form of the equation of this circle is . (simplify your answer. type an equation.) use the graphing tool to graph the circle. click to enlarge graph
Step1: Recall standard - form formula
The standard - form of the equation of a circle is $(x - h)^2+(y - k)^2=r^2$, where $(h,k)$ is the center and $r$ is the radius.
Step2: Substitute values
Given $h=-4$, $k = - 3$, and $r = 2$. Substitute these values into the standard - form formula:
$(x-(-4))^2+(y - (-3))^2=2^2$, which simplifies to $(x + 4)^2+(y + 3)^2=4$.
Step3: Expand to get general form
Expand $(x + 4)^2+(y + 3)^2=4$.
$(x + 4)^2=x^{2}+8x + 16$ and $(y + 3)^2=y^{2}+6y+9$.
So $x^{2}+8x + 16+y^{2}+6y + 9=4$.
Combine like - terms: $x^{2}+y^{2}+8x + 6y+16 + 9-4=0$.
The general form is $x^{2}+y^{2}+8x + 6y+21 = 0$.
Snap & solve any problem in the app
Get step-by-step solutions on Sovi AI
Photo-based solutions with guided steps
Explore more problems and detailed explanations
The standard form of the equation of this circle is $(x + 4)^2+(y + 3)^2=4$.
The general form of the equation of this circle is $x^{2}+y^{2}+8x + 6y+21 = 0$.