Test Index

UPSC NDA I 2019 Mathematics Solved Paper

Question : 60 of 120

Marks: +1, -0

The circle

x^{2} + y^{2} + 4x - 7y + 12 = 0

cut an intercept on y- axis equal

Solution:

Concept:

The intercept made by a circle on the y‑axis is given by the formula

2\sqrt{f^2 - c}

when the circle equation is

x^2 + y^2 + 2gx + 2fy + c = 0

Explanation:

The given circle equation is

x^2 + y^2 + 4x - 7y + 12 = 0

Compare it with the standard form

x^2 + y^2 + 2gx + 2fy + c = 0

We get

2g = 4

, so

g = 2

Also

2f = -7

, so

f = -\frac{7}{2}

The constant term is

c = 12

The y‑axis intercept length is

2\sqrt{f^2 - c}

Compute

f^2 = \left(-\frac{7}{2}\right)^2 = \frac{49}{4}

Then

f^2 - c = \frac{49}{4} - 12 = \frac{49}{4} - \frac{48}{4} = \frac{1}{4}

2\sqrt{f^2 - c} = 2 \times \sqrt{\frac{1}{4}} = 2 \times \frac{1}{2} = 1

Therefore the y‑axis intercept is 1 unit.

Answer:

1 (Option A)

Go to Question:

Prev Question

Next Question