The value of k such that the circles x^2 + y^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.For the first circle $x^2 + y^2 + kx + 4y + 2 = 0$,we have $g_1 = \frac{k}{2}$

Vedclass · Accepted Answer

$\frac{-10}{3}$

Vedclass · Answer

(C) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.For the first circle $x^2 + y^2 + kx + 4y + 2 = 0$,we have $g_1 = \frac{k}{2}$,$f_1 = 2$,and $c_1 = 2$.For the second circle,divide by $2$ to get $x^2 + y^2 - 2x - \frac{3}{2}y + \frac{k}{2} = 0$. Thus,$g_2 = -1$,$f_2 = -\frac{3}{4}$,and $c_2 = \frac{k}{2}$.The condition for two circles to intersect orthogonally is $2(g_1g_2 + f_1f_2) = c_1 + c_2$.Substituting the values: $2\left[\left(\frac{k}{2}\right)(-1) + (2)\left(-\frac{3}{4}\right)\right] = 2 + \frac{k}{2}$.$2\left[-\frac{k}{2} - \frac{3}{2}\right] = 2 + \frac{k}{2}$.$-k - 3 = 2 + \frac{k}{2}$.$-5 = k + \frac{k}{2} = \frac{3k}{2}$.$k = -\frac{10}{3}$.

The value of $k$ such that the circles $x^2 + y^2 + kx + 4y + 2 = 0$ and $2(x^2 + y^2) - 4x - 3y + k = 0$ cut orthogonally is:

Similar Questions

If the circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 10x + \lambda = 0$ touch externally,then $\lambda$ is equal to:

If the circles $x^2+y^2+2kx-4y+1=0$ and $x^2+y^2-8x-12y+43=0$ touch each other,then $k=$

The radical axis of the circles $S_1: x^2+y^2-4x+6y-10=0$ and $S_2: x^2+y^2+2x-6y+2=0$ cuts the circle $S_1$ in

The equation of the circle passing through $(0,0)$ and cutting orthogonally the circles $x^2+y^2+6x-15=0$ and $x^2+y^2-8y-10=0$ is

If one of the diameters of the circle $x^{2}+y^{2}-2x-6y+6=0$ is a chord of another circle $'C'$,whose center is at $(2,1)$,then its radius is..........

Vedclass Test Series

Exam Paper Generator

Online Exam Module