If the circles $x^{2} + y^{2} + 2 x + 2 k y + 6 = 0$ and $x^{2} + y^{2} + 2 k y + k = 0$ intersect orthogonally, then $k$ is

Answer (Detailed Solution Below)View answer ▾

Answer: Option A — $2$ or $- \frac{3}{2}$

Detailed Solution

First circle: $g_{1} = 1$ , $f_{1} = k$ , $c_{1} = 6$ . Second circle: $g_{2} = 0$ , $f_{2} = k$ , $c_{2} = k$ .

Orthogonality: $2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2}$ gives $2 k^{2} = 6 + k$ , i.e. $2 k^{2} - k - 6 = 0$ .

$(k - 2) (2 k + 3) = 0$ , so $k = 2$ or $k = - \frac{3}{2}$ .

✓ Check

k = 2

2 \cdot 4 = 8 = 6 + 2

k = - \frac{3}{2}

2 \cdot \frac{9}{4} = 4.5 = 6 - 1.5

⚠ Common trap

(D)

- 2

\frac{3}{2}

are the roots of

2 k^{2} + k - 6 = 0

, from a sign slip on the

k

term.

Quick tip

Orthogonal: $2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{2} \Rightarrow 2 k^{2} = 6 + k$

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Circle — Preparation

This question tests your understanding of Circle, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.

In coordinate geometry, a circle is the set of points at a fixed distance (the radius) from a fixed point (the centre). This topic covers the standard equation (x - h)^2 + (y - k)^2 = r^2, the general equation x^2 + y^2 + 2gx + 2fy + c = 0 with centre (-g, -f), the diameter form, and parametric equations. Exam questions also test the position of a point relative to a circle, the condition for a line to be a tangent, equations of tangents and normals, the length of a tangent from an external point, and the intersection of two circles. The fastest approach is often to compare the distance from the centre to a line with the radius: equal means tangent, smaller means the line cuts the circle. Practise the circle questions below, each with a step-by-step solution.

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If the circles $x^{2} + y^{2} + 2 x + 2 k y + 6 = 0$ and $x^{2} + y^{2} + 2 k y + k = 0$ intersect orthogonally, then $k$ is

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If the circles $x^{2} + y^{2} + 2 x + 2 k y + 6 = 0$ and $x^{2} + y^{2} + 2 k y + k = 0$ intersect orthogonally, then $k$ is