The equation of the circle which passes throu | vedclass

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

Question

(C) Let the required equation of the circle be $x^2+y^2+2gx+2fy=0$ (since it passes through the origin,$c=0$).For the circle $x^2+y^2+2g_1x+2f_1y+c_1=

Vedclass · Accepted Answer

$3x^2+3y^2-8x+29y=0$

Vedclass · Answer

(C) Let the required equation of the circle be $x^2+y^2+2gx+2fy=0$ (since it passes through the origin,$c=0$).For the circle $x^2+y^2+2g_1x+2f_1y+c_1=0$ to cut $x^2+y^2+2g_2x+2f_2y+c_2=0$ orthogonally,the condition is $2g_1g_2+2f_1f_2=c_1+c_2$.For the first circle $x^2+y^2-6x+8=0$,we have $g_1=-3, f_1=0, c_1=8$. Applying the condition: $2g(-3)+2f(0)=0+8$ $\Rightarrow -6g=8$ $\Rightarrow g=-\frac{4}{3}$.For the second circle $x^2+y^2-2x-2y-7=0$,we have $g_2=-1, f_2=-1, c_2=-7$. Applying the condition: $2g(-1)+2f(-1)=0-7 \Rightarrow -2g-2f=-7$.Substituting $g=-\frac{4}{3}$ into the second equation: $-2(-\frac{4}{3})-2f=-7$ $\Rightarrow \frac{8}{3}-2f=-7$ $\Rightarrow 2f = 7+\frac{8}{3} = \frac{29}{3}$ $\Rightarrow f=\frac{29}{6}$.The equation is $x^2+y^2+2(-\frac{4}{3})x+2(\frac{29}{6})y=0$.$x^2+y^2-\frac{8}{3}x+\frac{29}{3}y=0$.Multiplying by $3$,we get $3x^2+3y^2-8x+29y=0$.

The equation of the circle which passes through the origin and cuts orthogonally each of the circles $x^2+y^2-6x+8=0$ and $x^2+y^2-2x-2y-7=0$ is

Similar Questions

The circuit shown in the following figure contains two diodes $D_1$ and $D_2$, each with a forward resistance of $50 \, \Omega$ and infinite backward resistance. If the battery voltage is $6 \, V$, find the current through the $100 \, \Omega$ resistor (in amperes).

Which of the following is not a fat-soluble vitamin?

The number of atoms/molecules contained in one face-centred cubic $(FCC)$ unit cell of a monoatomic substance is

If $\int\limits_0^{\frac{\pi }{3}} {\frac{{\tan \theta }}{{\sqrt {2k \sec \theta } }}} \,d\theta = 1 - \frac{1}{\sqrt{2}}, (k > 0),$ then the value of $k$ is

Find the value of $V_{AB}$ in the given circuit.

Vedclass Test Series

Exam Paper Generator

Online Exam Module