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(x-h)^(2)+(y-k)^(2)=r^(2)
The formula relates the 
x - and 
y-coordinates of a point on a circle with a center at 
(h,k) to the radius 
r. Which of the following equations correctly gives 
y in terms of 
x,h,k, and 
r ?
Choose 1 answer:
(A) 
y=r-x+h+k
(B) 
y=r-(x-h)^(2)+k
(C) 
y=+-sqrt(r^(2)-(x-h)^(2))+k
(D) 
y=+-sqrt(r^(2)-(x-h)^(2)+k)
Show calculator

Google Classroom\newline(xh)2+(yk)2=r2 (x-h)^{2}+(y-k)^{2}=r^{2} \newlineThe formula relates the x x - and y y -coordinates of a point on a circle with a center at (h,k) (h, k) to the radius r r . Which of the following equations correctly gives y y in terms of x,h,k x, h, k , and r r ?\newlineChoose 11 answer:\newline(A) y=rx+h+k y=r-x+h+k \newline(B) y=r(xh)2+k y=r-(x-h)^{2}+k \newline(C) y=±r2(xh)2+k y= \pm \sqrt{r^{2}-(x-h)^{2}}+k \newline(D) y y 00\newlineShow calculator

Full solution

Q. Google Classroom\newline(xh)2+(yk)2=r2 (x-h)^{2}+(y-k)^{2}=r^{2} \newlineThe formula relates the x x - and y y -coordinates of a point on a circle with a center at (h,k) (h, k) to the radius r r . Which of the following equations correctly gives y y in terms of x,h,k x, h, k , and r r ?\newlineChoose 11 answer:\newline(A) y=rx+h+k y=r-x+h+k \newline(B) y=r(xh)2+k y=r-(x-h)^{2}+k \newline(C) y=±r2(xh)2+k y= \pm \sqrt{r^{2}-(x-h)^{2}}+k \newline(D) y y 00\newlineShow calculator
  1. Start Equation Circle: Start with the given equation of a circle.\newlineThe equation of a circle with center (h,k)(h, k) and radius rr is given by:\newline(xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2\newlineWe need to solve this equation for yy in terms of xx, hh, kk, and rr.
  2. Isolate Term: Isolate the (yk)2(y - k)^2 term.\newlineTo solve for yy, we first need to isolate the (yk)2(y - k)^2 term on one side of the equation. We do this by subtracting (xh)2(x - h)^2 from both sides of the equation:\newline(yk)2=r2(xh)2(y - k)^2 = r^2 - (x - h)^2
  3. Take Square Root: Take the square root of both sides.\newlineTo solve for yy, we take the square root of both sides of the equation. Remember that taking the square root of both sides introduces a plus or minus (±\pm) sign:\newlineyk=±r2(xh)2y - k = \pm\sqrt{r^2 - (x - h)^2}
  4. Solve for y: Solve for y.\newlineFinally, we add kk to both sides of the equation to solve for y:\newliney=k±r2(xh)2y = k \pm \sqrt{r^2 - (x - h)^2}

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