If the equation of the circle which passes th | vedclass

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

Question

(D) Let the required circle be $S_3: x^2+y^2+2gx+2fy+c=0$. Given circles are $S_1: x^2+y^2-4x-6y+4=0$ and $S_2: x^2+y^2+6x-4y+15=0$. Since $S_3$ cuts

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) Let the required circle be $S_3: x^2+y^2+2gx+2fy+c=0$. Given circles are $S_1: x^2+y^2-4x-6y+4=0$ and $S_2: x^2+y^2+6x-4y+15=0$. Since $S_3$ cuts $S_1$ orthogonally,$2g(-2) + 2f(-3) = c + 4 \implies -4g - 6f = c + 4 \implies 4g + 6f + c = -4 \dots(1)$. Since $S_3$ cuts $S_2$ orthogonally,$2g(3) + 2f(-2) = c + 15 \implies 6g - 4f = c + 15 \implies 6g - 4f - c = 15 \dots(2)$. Since $S_3$ passes through $(1,1)$,$1^2 + 1^2 + 2g(1) + 2f(1) + c = 0 \implies 2g + 2f + c = -2 \dots(3)$. Subtracting $(3)$ from $(1)$: $(4g + 6f + c) - (2g + 2f + c) = -4 - (-2) \implies 2g + 4f = -2 \implies g + 2f = -1 \implies g = -1 - 2f$. Adding $(1)$ and $(2)$: $(4g + 6f + c) + (6g - 4f - c) = -4 + 15 \implies 10g + 2f = 11$. Substituting $g = -1 - 2f$ into $10g + 2f = 11$: $10(-1 - 2f) + 2f = 11 \implies -10 - 20f + 2f = 11 \implies -18f = 21 \implies f = -\frac{7}{6}$. Then $g = -1 - 2(-\frac{7}{6}) = -1 + \frac{7}{3} = \frac{4}{3}$. From $(3)$,$c = -2 - 2g - 2f = -2 - 2(\frac{4}{3}) - 2(-\frac{7}{6}) = -2 - \frac{8}{3} + \frac{7}{3} = -2 - \frac{1}{3} = -\frac{7}{3}$. Finally,$5g + 2f + c = 5(\frac{4}{3}) + 2(-\frac{7}{6}) + (-\frac{7}{3}) = \frac{20}{3} - \frac{7}{3} - \frac{7}{3} = \frac{6}{3} = 2$.

If the equation of the circle which passes through the point $(1,1)$ and cuts both the circles $x^2+y^2-4x-6y+4=0$ and $x^2+y^2+6x-4y+15=0$ orthogonally is $x^2+y^2+2gx+2fy+c=0$,then $5g+2f+c=$

Similar Questions

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

The line $x-2=0$ cuts the circle $x^2+y^2-8x-2y+8=0$ at $A$ and $B$. The equation of the circle passing through the points $A$ and $B$ and having the least radius is

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

The radical centre of the circles $x^2+y^2+2x+3y+1=0$,$x^2+y^2+x-y+3=0$,and $x^2+y^2-3x+2y+5=0$ is

The number of common tangents to the circles $x^2 + y^2 - 8x - 2y + 1 = 0$ and $x^2 + y^2 + 6x + y = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module