The circle $x^{2} + y^{2} = 9$ is contained in the circle $x^{2} + y^{2} - 6 x - 8 y + 25 = c^{2}$ if

Answer (Detailed Solution Below)View answer ▾

Answer: Option D — $c = 10$

Detailed Solution

$x^{2} + y^{2} - 6 x - 8 y + 25 = c^{2}$ is $(x - 3)^{2} + (y - 4)^{2} = c^{2}$ : centre $(3, 4)$ , radius $c$ .

The circle $x^{2} + y^{2} = 9$ has centre $(0, 0)$ and radius $3$ ; the distance between the centres is $9 + 16 = 5$ .

The small circle lies inside when $c \geq 5 + 3 = 8$ . Among the options only $c = 10$ works.

✓ Check

With

c = 10

, the point of the small circle farthest from

(3, 4)

is at distance

5 + 3 = 8 < 10

, so the whole small circle is inside.

⚠ Common trap

(C)

c = 5

only makes the big circle pass through

(0, 0)

, the centre of the small circle; half of the small circle is still outside.

Quick tip

$r_{2} \geq d + r_{1} = 5 + 3 = 8$

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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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