Question

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

Answer 1

equations of circles are

${x}^{2} + {y}^{2} + 2x + 2ky + 6 = 0 \\ here \: \: g \frac{}{1} = - 1 \: \: \: \: f \frac{}{1} = - k \: \: \: \: c \frac{}{1} = 6 \\ \\ other \: eq. \: is \\ {x}^{2} + {y}^{2} + 2ky + k = 0 \\ here \: \: g \frac{}{2} = 0 \: \: \: \: f \frac{}{2} = - k \: \: \: \: c \frac{}{2} = k \\ \\ for \: the \: circles \: to \: intersect \: each \\ other \: orthogonally \\ 2(g \frac{}{1} )(g \frac{}{2} ) + 2(f \frac{}{1} )(f \frac{}{2} ) = c \frac{}{1} \: \: + \: \: c \frac{}{2} \\ 2( - 1)(0) + 2( - k)( - k) = 6 + k \\ 2 {k }^{2} - k - 6 = 0 \\ {2k}^{2} - 4k + 3k - 6 = 0 \\ 2k(k - 2) + 3(k - 2) = 0 \\ (k - 2)(2k + 3) = 0 \\ therefore \\ k = 2 \: \: \: or \: \: \: \frac{ - 3}{2}$

hope ot helps（＾＿－）≡★

Answer 2

The circles $x^{2} +y^{2}+2x+2ky+6=0$  and $x^{2} +y^2+2ky+k=0$ intersect orthogonally then $k$ equals to $2$ or $\frac{-3}{2}$.

Step-by-step explanation:

Given:

The circles x2 + y2 +2x + 2ky + 6 = 0 and x2 + y2 + 2ky + k = 0 intersect orthogonally.

Let $R_{1}$ and $R_{2}$ be the radius of both circles and $d$ be the distance between the centres of two circles

Two circles are intersect orthogonally if $R_{1}^{2}+R_{2}^{2}=d^{2}$

The equations of circles are

$x^{2} +y^{2}+2x+2ky+6=0$

On arranging the given equation, we

$x^{2} +2x+4+y^{2}+2ky+2 =0$

Adding $k^{2}$ on both the side

$x^{2} +2x+4+y^{2}+2ky+2+k^{2} =0+k^{2}$

$(x^{2} +2x+4)+(y^{2}+2ky+k^{2})+2 =k^{2}$

$(x+2)^{2}+(y+k)^{2}+2=k^2$

$(x+2)^{2}+(y+k)^{2}=k^2-2$                    $(1)$

Now solving second equation

$x^{2} +y^2+2ky+k=0$

Adding $k^{2}$ on both the side

$x^{2} +y^2+2ky+k+k^2=k^2$

$x^{2} +(y+k)^{2}+k=k^2$

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

Both the equations are converted to general form $(x-h)^2+(y-k)^2=r^2$, where (h,k) is centre of circle.

Hence centres of circles are $C_1(-2,-k)$ and $C_2(0,-k)$

Apply the condition of orthogonal we get,

$R_1^2+R_2^2=d^2$

$k^2-2+k^2-k$ = $(0-(-2))^2+(-k-(-k))^2$

$2k^2-k-2$ = $4^2+(-k+k)^2$

                              = $4$

$2k^2-k$ = $4+2$

                           = $6$

$2k^2-k-6$ = $0$

$2k^2-4k+3k-6=0$

$2k(k-2)+3(k-2)=0$

$(k-2)(2k+3)=0$

Hence

$k-2=0$ , $2k+3=0$

$k=2$ , $2k=-3$

                        $k=\frac{-3}{2}$