What is the radius of the circle 4 x²+4 y²-20 x+12 y-15=0 ? [NDA-I 2021]

Correct Answer is : 3.53.53.5 units

Correct Answer is : 3.53.53.5 units

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UPSC NDA I 2021 Mathematics Solved Paper

Question : 11 of 120

Marks: +1, -0

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

Solution: 👈: Video Solution

Concept

Equation of circle having centre

(h, k)

and radius

r

(x-h)^{2}+(y-k)^{2}=r^{2}

Calculation:

Given equation of circle is,

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

\Rightarrow 4 x^{2}-20 x+25+4 y^{2}+12 y+9-15-34=0

\Rightarrow (2 x-5)^{2}+(2 y+3)^{2}=49

\Rightarrow 4\left(x-\;\frac{5}{2}\right)^{2}+4\left(y+\;\frac{3}{2}\right)^{2}=49

\Rightarrow \left(x-\;\frac{5}{2}\right)^{2}+\left(y+\;\frac{3}{2}\right)^{2}=\left(\;\frac{7}{2}\right)^{2}

So, the radius of the circle

=\;\frac{7}{2}

units

=3.5

units

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