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The two circles, which pass through $\mathrm{(0, a) \: and \: (0,-a)}$ and touch the line $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ , will cut orthogonally if $\mathrm{c^2=a^2\left(k+m^2\right)}$ where $\mathrm{ k=}$

Option: 1
$1$

Option: 2
$2$

Option: 3
$3$

Option: 4
$4$

Answers (1)

best_answer

Let the equation of the circles be

$\mathrm{ x^2+y^2+2 g x+2 f y+d=0 }$ ........(i)

Since these circles pass through $\mathrm{ (0, a) }$ and $\mathrm{ (0,-a) }$ then

$\mathrm{ a^2+2 f a+d=0 }$ .............(ii)
and $\mathrm{ a^2-2 f a+d=0 }$ .............(iii)

Solving (ii) and (iii), we get $\mathrm{ f=0(a \neq 0) \: and\: \: d=-a^2 }$ . Substituting these values of $\mathrm{ f \: and \: d }$ in (i) we obtain

$\mathrm{ x^2+y^2+2 g x-a^2=0 }$ ...........(iv)

Now $\mathrm{ y=m x+c }$ touch this circle, therefore, length of the perpendicular from the center $\mathrm{ = }$ radius

$\mathrm{ \frac{|-m g-0+c|}{\sqrt{1+m^2}}=\sqrt{\left(g^2+a^2\right)} }$

$\mathrm{ (c-m g)^2=\left(1+m^2\right)\left(g^2+a^2\right) }$

$\mathrm{ \text { or } g^2+2 m c g+a^2\left(1+m^2\right)-c^2 }$

Let $\mathrm{ \mathrm{g}_1, \mathrm{~g}_2 }$ are the roots of this equation

$\therefore \quad \mathrm{g}_1 \mathrm{~g}_2=\mathrm{a}^2\left(1+\mathrm{m}^2\right)-\mathrm{c}^2$ ............(v)

Now, the equation of the two circles represented by (iv) are

$\mathrm{x^2+y^2+2 g_1 x-a^2=0}$ and $\mathrm{x^2+y^2+2 g_2 x-a^2=0}$ . These two circles will be orthogonal if

$\mathrm{ 2 g_1 g_2+0=-a^2-a^2 }$

or $\mathrm{ g_1 g_2=-a^2}$

$\mathrm{g}_1 \mathrm{~g}_2=-\mathrm{a}^2$ ............(vi)

From (v) and (vi)

$\mathrm{ -a^2=a^2\left(1+m^2\right)-c^2 }$
or
$\mathrm{ c^2=a^2\left(2+m^2\right) }$

which is the required condition.

Hence option 2 is correct.

Posted by

Divya Prakash Singh

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