The radical centre of the system of circles, and x²+y²+4x+7=0 2(x²+y²)+3x+5y+9=0 x²+y²+y=0 { is }

Correct Answer is : (−2,−1)(-2,-1)(−2,−1)

Correct Answer is : (−2,−1)(-2,-1)(−2,−1)

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CGPET 2014 Solved Paper

Section: Mathematics

Question : 134 of 150

Marks: +1, -0

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

2(x^2+y^2)+3x+5y+9=0

x^2+y^2+y=0 \text{ is }

Solution:

Given system of circles is

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

. . . (i)

2(x^2+y^2)+3x+5y+9=0

. . . (ii)

\text{ or } x^2+y^2+\frac{3}{2}x+\frac{5}{2}y+\frac{9}{2}=0

. . . (iii)

\text{ and } x^2+y^2+y=0

The radical centre can be obtained by solving the Eqs. (i), (ii) and (iii).

On subtracting Eq. (ii) from Eq. (i), we get

4x-\frac{3}{2}x-\frac{5}{2}y+7-\frac{9}{2}=0

\Rightarrow \frac{5}{2}x-\frac{5}{2}y+\frac{5}{2}=0

x-y+1=0

. . . (iv)

On subtracting Eq. (iii) from Eq. (i), we get

4x-y+7=0

. . . (v)

On solving Eqs. (iv) and (v), we get and

y=-1

Hence, radical centre is

(-2,-1)

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