Question

Circles are drawn through the point \( (2, 0) \) to cut intercepts of length 5 units on the X-axis. If their centre lies in the first quadrant, then their equation is

• A. \( 3x^2 + 3y^2 - 27x - 2ky + 42 = 0, \; k \in \mathbb{R}^+ \)
• B. \( x^2 + y^2 - 2kx - 9y + 14 = 0, \; k \in \mathbb{R}^+ \)
• C. \( x^2 + y^2 - 9x - 2ky + 14 = 0, \; k \in \mathbb{R}^+ \)
• D. \( x^2 + y^2 - 9x - 2ky - 42 = 0, \; k \in \mathbb{R}^+ \)

Hint:

When a circle cuts a chord of a known length on the X-axis, use perpendicular distance and Pythagoras theorem to relate radius and center.

Answer 1

The Correct Option is C

Step 1: Understand the geometric conditions. The circle must: - Pass through \( (2, 0) \) - Intersect the X-axis with a chord of length 5 units - Have its center in the first quadrant Let the general equation of a circle be: \[ x^2 + y^2 + Dx + Ey + F = 0 \] Step 2: Use the chord length condition. Let the center be \( (h, k) \). Then the radius \( r = \sqrt{h^2 + k^2} \). The chord of length 5 implies that the perpendicular from the center to X-axis (which is \( k \)) divides the chord into two equal parts: \[ \text{Half chord length} = \sqrt{r^2 - k^2} = \frac{5}{2} \Rightarrow r^2 - k^2 = \left(\frac{5}{2}\right)^2 = \frac{25}{4} \] Step 3: Apply point condition. The point \( (2, 0) \) lies on the circle. Substituting into the general form gives: \[ % Option (2)^2 + (0)^2 - 9(2) - 2k(0) + 14 = 0 \Rightarrow 4 - 18 + 14 = 0 \Rightarrow 0 = 0 \] So the point satisfies the equation: \[ x^2 + y^2 - 9x - 2ky + 14 = 0,
k \in \mathbb{R}^+ \]