Question

The one of possible real value of ‘m’ for which the circles, x2 + y2 + 4x + 2(m2 + m) y + 6 = 0 and x2 + y2 + (2y + 1)(m2 + m) = 0 intersect orthogonally is
1) 3
2) -1
3) 2
4) -2​

Answer 1

The one of possible real value of m is -2

Therefore, option (4) is correct.

Explanation:

For circle 1

$x^2+y^2+4x+2(m^2+m)y+6=0$

Comparing this with general equation of circle

$x^2+y^2+2g_1x+2f_1y+c_1=0$  ....... (1)

$g_1=2$

$f_1=m^2+m$

$c_1=6$

Similarly, for circle 2

$x^2+y^2+(2y+1)(m^2+m)=0$

$\implies x^2+y^2+2(m^2+m)y+(m^2+m)=0$

Comparing this with general equation of circle

$x^2+y^2+2g_2x+2f_2y+c_2=0$ ......... (2)

$g_2=0$

$f_2=m^2+m$

$c_2=m^2+m$

We know that the condition that two circles (1) and (2) intersect orthogonally is

$\boxed{2g_1g_2+2f_1f_2=c_1+c_2}$

Thus,

$2\times 2\times 0+2*(m^2+m)(m^2+m)=6+(m^2+m)$

$\implies 2(m^2+m)^2-(m^2+m)-6=0$

Let $m^2+m=t$

$2t^2-t-6=0$

$\implies 2t^2-4t+3t-6=0$

$\implies 2t(t-2)+3(t-2)=0$

$\implies (t-2)(2t+3)=0$

$\implies t=2, -\frac{3}{2}$

When $t=-3/2$

$m^2+m=-\frac{3}{2}$

$\implies 2m^2+2m+3=0$

This quadratic equation does not have a real root

Again, taking

$m^2+m=2$

$\implies m^2+m-2=0$

$\implies m^2+2m-m-2=0$

$\implies m(m+2)-1(m+2)=0$

$\implies (m+2)(m-1)=0$

$\implies m=-2,1$

Therefore, the one possible real value of m for which the given circles intersect orthogonally is -2

Hope this answer is helpful.

Know More:

Q: If the circles x2 + y2 +2x + 2ky + 6 = 0 and x2 + y2 + 2ky + k = 0 intersect orthogonally then k equals ?

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