Question

The equation of the circle whose radius is $\sqrt{7}$ and concentric with the circle $x^{2}+y^{2}-8x+6y-11=0$ is

• A. $x^{2}+y^{2}-8x+6y+7=0$
• B. $x^{2}+y^{2}-8x+6y+18=0$
• C. $x^{2}+y^{2}-8x+6y-4=0$
• D. $x^{2}+y^{2}-8x+6y-18=0$
• E. $x^{2}+y^{2}-8x+6y-7=0$

Hint:

Geometry Tip: Concentric circles always have the exact same $x^2, y^2, x,$ and $y$ terms. You only ever need to calculate the new constant at the end based on the new radius!

Answer 1

The Correct Option is B

Concept:
Two circles are "concentric" if they share the exact same centre point. Therefore, their equations will have identical $x$ and $y$ terms (like $x^2 + y^2 + 2gx + 2fy$) and will only differ by their constant term $C$. We can use the given radius to solve for this new constant.

Step 1: Find the centre of the original circle.
The original circle is $x^2 + y^2 - 8x + 6y - 11 = 0$. Find $g$ and $f$: $2g = -8 \implies g = -4$, and $2f = 6 \implies f = 3$. The shared centre is $(-g, -f) = (4, -3)$.

Step 2: Set up the equation for the new concentric circle.
Since it has the same centre, its equation starts the exact same way: $$x^2 + y^2 - 8x + 6y + C = 0$$ We just need to determine the new constant $C$.

Step 3: Apply the radius formula to the new circle.
We are given that the new circle has a radius of $r = \sqrt{7}$. Using the formula $r = \sqrt{g^2 + f^2 - C}$: $$\sqrt{7} = \sqrt{(-4)^2 + (3)^2 - C}$$

Step 4: Solve for the constant C.
Square both sides of the equation to remove the square roots: $$7 = 16 + 9 - C$$ $$7 = 25 - C$$ $$C = 25 - 7 = 18$$

Step 5: Construct the final equation.
Substitute $C = 18$ back into the general equation from
Step 2: $$x^2 + y^2 - 8x + 6y + 18 = 0$$ Hence the correct answer is (B) $x^{2+y^{2}-8x+6y+18=0$}.