Question

The equation of a circle whose Centre is (2, -1) and which passes through the point (3, 6) is

• A. $x^{2} + y^{2} + 4x + 2y - 45 = 0$
• B. $x^{2} + y^{2} - 2x + 2y - 50 = 0$
• C. $x^{2} + y^{2} + 2x + 2y - 50 = 0$
• D. $x^{2} + y^{2} - 4x + 2y - 45 = 0$

Hint:

If the center is $(h, k)$, the linear terms in the general form are always $-2hx$ and $-2ky$. For $(2, -1)$, look for $-4x$ and $+2y$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The equation of a circle with centre \((h, k)\) and radius \(r\) is given by \((x-h)^2 + (y-k)^2 = r^2\). Since the circle passes through a specific point, the distance between the centre and that point equals the radius.
Step 2: Key Formula or Approach:
1. Distance formula: \(r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
2. Standard circle equation: \((x - h)^2 + (y - k)^2 = r^2\)
Step 3: Detailed Explanation:
First, find the radius \(r\) using the distance between the centre \((2, -1)\) and the point \((3, 6)\): \[ r^2 = (3 - 2)^2 + (6 - (-1))^2 \] \[ r^2 = (1)^2 + (7)^2 = 1 + 49 = 50 \] Now, substitute the centre \((2, -1)\) and \(r^2 = 50\) into the circle equation: \[ (x - 2)^2 + (y - (-1))^2 = 50 \] \[ (x - 2)^2 + (y + 1)^2 = 50 \] Expand the equation: \[ (x^2 - 4x + 4) + (y^2 + 2y + 1) = 50 \] \[ x^2 + y^2 - 4x + 2y + 5 - 50 = 0 \] \[ x^2 + y^2 - 4x + 2y - 45 = 0 \]
Step 4: Final Answer:
The equation of the circle is \( x^2 + y^2 - 4x + 2y - 45 = 0 \).