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UPSC NDA I 2021 Mathematics Solved Paper

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Question : 11 of 120

Marks: +1, -0

What is the radius of the circle

4x^{2} + 4y^{2} - 20x + 12y - 15 = 0

?

14 units
10.5 units
7 units
3.5 units

Solution:

Concept:

The equation of a circle in general form

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

has centre

(-g, -f)

and radius

\sqrt{g^2 + f^2 - c}

.

Explanation:

Start with the given equation:

4x^2 + 4y^2 - 20x + 12y - 15 = 0

.

Divide the whole equation by

4

to get:

x^2 + y^2 - 5x + 3y - \frac{15}{4} = 0

.

Compare this with the general form

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

.

We get

2g = -5

, so

g = -\frac{5}{2}

.

Also

2f = 3

, so

f = \frac{3}{2}

.

And

c = -\frac{15}{4}

.

Now, radius

= \sqrt{g^2 + f^2 - c} = \sqrt{\left(-\frac{5}{2}\right)^2 + \left(\frac{3}{2}\right)^2 - \left(-\frac{15}{4}\right)}

.

Compute:

\sqrt{\frac{25}{4} + \frac{9}{4} + \frac{15}{4}} = \sqrt{\frac{25+9+15}{4}} = \sqrt{\frac{49}{4}} = \frac{7}{2}

units.

Answer:

Radius

= 3.5

units.

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