Question

The centre and radius of the circle $x^2 + y^2 - 4x - 8y - 41 = 0$ are

• A. $(1, -2), 5$
• B. $(2, 1), 3$
• C. $(2, 4), \sqrt{61}$
• D. $(1, -2), \sqrt{51}$

Hint:

A quick way to find the centre is to take the coefficients of $x$ and $y$, divide them by $-2$. Here: $-4/(-2) = 2$ and $-8/(-2) = 4$. This gives you the centre $(2, 4)$ instantly!

Answer 1

The Correct Option is C

Step 1: General Equation Comparison: The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. Given equation: $x^2 + y^2 - 4x - 8y - 41 = 0$. Comparing the coefficients: $2g = -4 \implies g = -2$ $2f = -8 \implies f = -4$ $c = -41$

Step 2: Determine the Centre: The centre of the circle is given by $(-g, -f)$. $$\text{Centre} = (-(-2), -(-4)) = (2, 4)$$

Step 3: Calculate the Radius ($r$): The formula for the radius is $r = \sqrt{g^2 + f^2 - c}$. Substituting the values: $$r = \sqrt{(-2)^2 + (-4)^2 - (-41)}$$ $$r = \sqrt{4 + 16 + 41}$$ $$r = \sqrt{61}$$ Therefore, the centre is $(2, 4)$ and the radius is $\sqrt{61}$.