If x^2 + y^2 + px + 3y - 5 = 0 and x^2 + y^2 | vedclass

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

Question

(A) The condition 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 is $2g_1g_2 + 2f

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) The condition 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 is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Comparing the given equations with the standard form:For the first circle: $2g_1 = p \Rightarrow g_1 = \frac{p}{2}$,$2f_1 = 3 \Rightarrow f_1 = \frac{3}{2}$,$c_1 = -5$.For the second circle: $2g_2 = 5 \Rightarrow g_2 = \frac{5}{2}$,$2f_2 = p \Rightarrow f_2 = \frac{p}{2}$,$c_2 = 7$.Substituting these values into the condition:$2\left(\frac{p}{2}\right)\left(\frac{5}{2}\right) + 2\left(\frac{3}{2}\right)\left(\frac{p}{2}\right) = -5 + 7$$\frac{5p}{2} + \frac{3p}{2} = 2$$\frac{8p}{2} = 2$$4p = 2$$p = \frac{2}{4} = \frac{1}{2}$.

If $x^2 + y^2 + px + 3y - 5 = 0$ and $x^2 + y^2 + 5x + py + 7 = 0$ cut orthogonally,then $p$ is

Similar Questions

Two circles $S_1 = x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $S_2 = x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ cut each other orthogonally,then:

$A$ circle passes through the origin and has its centre on $y = x$. If it cuts ${x^2} + {y^2} - 4x - 6y + 10 = 0$ orthogonally,then the equation of the circle is

$\left(0, \frac{3}{4}\right)$ is the radical centre of the circles $S_1: x^2+y^2-2x+6y=0$,$S_2: x^2+y^2+2gx-2y+6=0$,and $S_3: x^2+y^2-12x+2fy+3=0$. If $S_2$ and $S_3$ intersect orthogonally,then $(g, f) =$

The equation of the circle passing through the point $(1, 1)$ and the points of intersection of $x^{2}+y^{2}-6x-8=0$ and $x^{2}+y^{2}-6=0$ is

The radical centre of the circles $x^2+y^2-4x-6y+5=0$,$x^2+y^2-2x-4y-1=0$ and $x^2+y^2-6x-2y=0$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module