Question

The normal at the point (3, 4) on a circle cuts the circle at the point (-1, -2). Then the equation of the circle is

• A. \(x^2 + y^2 + 2x - 2y - 13 = 0\)
• B. \(x^2 + y^2 - 2x - 2y - 11 = 0\)
• C. \(x^2 + y^2 - 2x + 2y + 12 = 0\)
• D. \(x^2 + y^2 - 2x - 2y + 14 = 0\)

Hint:

Diameter ends give circle equation: \((x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0\).

Answer 1

The Correct Option is B

To find the equation of the circle, we start by using the given information that the normal at the point \( (3, 4) \) on the circle cuts the circle again at the point \( (-1, -2) \). This implies that the points \( (3, 4) \) and \( (-1, -2) \) are diametrically opposite points on the circle, meaning the line through these points is a diameter of the circle.

Step 1: Find the center of the circle.

The midpoint of the diameter gives the center of the circle.

\(\text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{3 + (-1)}{2}, \frac{4 + (-2)}{2} \right) = (1, 1)\)

Step 2: Find the radius of the circle.

The radius is the distance from the center \( (1, 1) \) to one of the points, say \( (3, 4) \).

\(\text{Radius} = \sqrt{(3 - 1)^2 + (4 - 1)^2} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}\)

Step 3: Write the equation of the circle using the center and radius.

The general equation of a circle with center \( (h, k) \) and radius \( r \) is:

\((x - h)^2 + (y - k)^2 = r^2\)

Substitute \( h = 1 \), \( k = 1 \), and \( r^2 = 13 \):

\((x - 1)^2 + (y - 1)^2 = 13\)

Expanding the equation:

\((x^2 - 2x + 1) + (y^2 - 2y + 1) = 13\)

Simplifying:

\(x^2 + y^2 - 2x - 2y + 2 = 13\)\(x^2 + y^2 - 2x - 2y - 11 = 0\)

The equation of the circle is \(x^2 + y^2 - 2x - 2y - 11 = 0\), which matches the given option:

\(x^2 + y^2 - 2x - 2y - 11 = 0\)