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(x+3)^(2)+(y-4)^(2)=9 A circle in the xy-plane has the equation shown. If the x-coordinate of a point on the circle is -3 , what is a possible corresponding y coordinate?

(x+3)2+(y−4)2=9 (x+3)^{2}+(y-4)^{2}=9 \newlineA circle in the xy x y -plane has the equation shown. If the x x -coordinate of a point on the circle is −3-3 , what is a possible corresponding y y coordinate?

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Q. (x+3)2+(y−4)2=9 (x+3)^{2}+(y-4)^{2}=9 \newlineA circle in the xy x y -plane has the equation shown. If the x x -coordinate of a point on the circle is −3-3 , what is a possible corresponding y y coordinate?
  1. Substitute and Solve: We are given the equation of a circle (x+3)2+(y−4)2=9(x+3)^2 + (y-4)^2 = 9 and we need to find the y-coordinate when x=−3x = -3. First, let's substitute x=−3x = -3 into the equation and solve for yy. (−3+3)2+(y−4)2=9(-3 + 3)^2 + (y - 4)^2 = 9
  2. Simplify Left Side: Now, we simplify the left side of the equation.\newline(0)2+(y−4)2=9(0)^2 + (y - 4)^2 = 9\newline0+(y−4)2=90 + (y - 4)^2 = 9\newline(y−4)2=9(y - 4)^2 = 9
  3. Take Square Root: Next, we take the square root of both sides of the equation to solve for y−4y - 4.(y−4)2=±9\sqrt{(y - 4)^2} = \pm\sqrt{9}y−4=±3y - 4 = \pm3
  4. Find Solutions: We have two possible solutions for yy. Let's find both.\newlineFirst solution: y−4=3y - 4 = 3\newliney=3+4y = 3 + 4\newliney=7y = 7
  5. First Solution: Second solution: y−4=−3y - 4 = -3\newliney=−3+4y = -3 + 4\newliney=1y = 1

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