If the two circles 2x^2 + 2y^2 - 3x + 6y + k | vedclass

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

Question

(C) The given equations of the circles are:$2x^2 + 2y^2 - 3x + 6y + k = 0$Dividing by $2$,we get:$x^2 + y^2 - \frac{3}{2}x + 3y + \frac{k}{2} = 0$ ...

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) The given equations of the circles are:$2x^2 + 2y^2 - 3x + 6y + k = 0$Dividing by $2$,we get:$x^2 + y^2 - \frac{3}{2}x + 3y + \frac{k}{2} = 0$ .....$(i)$And the second circle is:$x^2 + y^2 - 4x + 10y + 16 = 0$ .....$(ii)$For two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ to cut orthogonally,the condition is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Comparing $(i)$ and $(ii)$ with the standard form:$g_1 = -\frac{3}{4}, f_1 = \frac{3}{2}, c_1 = \frac{k}{2}$$g_2 = -2, f_2 = 5, c_2 = 16$Substituting these values into the condition:$2(-\frac{3}{4})(-2) + 2(\frac{3}{2})(5) = \frac{k}{2} + 16$$3 + 15 = \frac{k}{2} + 16$$18 = \frac{k}{2} + 16$$\frac{k}{2} = 2$$k = 4$

If the two circles $2x^2 + 2y^2 - 3x + 6y + k = 0$ and $x^2 + y^2 - 4x + 10y + 16 = 0$ cut orthogonally,then the value of $k$ is

Similar Questions

Which of the following hydrides is electron-deficient?

The vector sum of two forces is perpendicular to their vector difference. In that case,the forces

If $f(x)=(\cos x)(\cos 2 x) \ldots(\cos n x)$,then $f^{\prime}(x)+\sum_{r=1}^n(r \tan r x) f(x)$ is equal to

If $\Delta$ is the area of the triangle formed by the positive $x$-axis and the normal and tangent to the circle $x^2+y^2=4$ at $(1, \sqrt{3})$,then $\Delta$ is equal to

The oxidation state of $Xe$ in $XeO_3$ and the bond angle in it respectively are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module