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 equation of the circle be ${x^2} + {y^2} + 2gx + 2fy = 0$ since it passes through the origin $(0, 0)$.Since the centre $(-g, -f)$ lies on

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the equation of the circle be ${x^2} + {y^2} + 2gx + 2fy = 0$ since it passes through the origin $(0, 0)$.Since the centre $(-g, -f)$ lies on the line $x + y = 4$,we have $-g - f = 4$,or $g + f = -4$ ... $(i)$.The circle cuts ${x^2} + {y^2} - 4x + 2y + 4 = 0$ orthogonally. The condition for orthogonality is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Here,$g_1 = g, f_1 = f, c_1 = 0$ and $g_2 = -2, f_2 = 1, c_2 = 4$.So,$2(g)(-2) + 2(f)(1) = 0 + 4$,which simplifies to $-4g + 2f = 4$,or $-2g + f = 2$ ... $(ii)$.Subtracting $(ii)$ from $(i)$: $(g + f) - (-2g + f) = -4 - 2$,which gives $3g = -6$,so $g = -2$.Substituting $g = -2$ into $(i)$: $-2 + f = -4$,so $f = -2$.The equation of the circle is ${x^2} + {y^2} + 2(-2)x + 2(-2)y = 0$,which is ${x^2} + {y^2} - 4x - 4y = 0$.

The equation of the circle which passes through the origin,has its centre on the line $x + y = 4$ and cuts the circle ${x^2} + {y^2} - 4x + 2y + 4 = 0$ orthogonally,is

Similar Questions

$\left(0, \frac{3}{4}\right)$ 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 $(g, f) =$

$P, Q$ and $R$ are the centres and $r_1, r_2, r_3$ are the radii respectively of three co-axial circles. Then $QRr_1^2 + RP r_2^2 + PQ r_3^2$ is equal to

The equation of the circle passing through the points of intersection of $x^2 + y^2 - 1 = 0$ and $x^2 + y^2 - 2x - 4y + 1 = 0$ and touching the line $x + 2y = 0$ is

Find the equation of the radical axis of the two circles $x^2 + y^2 - x + 1 = 0$ and $3(x^2 + y^2) + y - 1 = 0$.

If the curves $x^{2}-6x+y^{2}+8=0$ and $x^{2}-8y+y^{2}+16-k=0$ $(k>0)$ touch each other at a point,then the largest value of $k$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module