Question

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

• A. (2,-1) and 5
• B. (4, 2) and $\sqrt{20}$
• C. $(2,-1)$ and $\sqrt{5}$
• D. (-2, 1) and 5
• E. $(-2,1)$ and $\sqrt{5}$

Hint:

Geometry Tip: To quickly find the centre of a circle in general form, just take the coefficients of $x$ and $y$, divide them by 2, and flip their signs!

Answer 1

The Correct Option is C

Concept:
The standard equation of a general circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. The centre of this circle is given by the coordinates $(-g, -f)$ and its radius is calculated using the formula $r = \sqrt{g^2 + f^2 - c}$.

Step 1: Identify the coefficients from the given equation.
The given equation is $x^2 + y^2 - 4x + 2y = 0$. By comparing this with the standard equation $x^2 + y^2 + 2gx + 2fy + c = 0$, we extract: $2g = -4$ $2f = 2$ $c = 0$

Step 2: Solve for g and f.
Divide by 2 to find the values of $g$ and $f$: $g = -2$ $f = 1$

Step 3: Determine the centre of the circle.
The centre of the circle is located at $(-g, -f)$. Substitute the values we just found: $$\text{Centre} = (-(-2), -(1)) = (2, -1)$$

Step 4: Calculate the radius.
Plug $g$, $f$, and $c$ into the radius formula $r = \sqrt{g^2 + f^2 - c}$: $$r = \sqrt{(-2)^2 + (1)^2 - 0}$$ $$r = \sqrt{4 + 1}$$ $$r = \sqrt{5}$$

Step 5: Match with the given options.
The centre is $(2, -1)$ and the radius is $\sqrt{5}$. This exactly matches option (C). Hence the correct answer is (C) $(2,-1)$ and $\sqrt{5$}.