If (alpha, beta) is the centre of the circle | vedclass

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

Question

(D) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$. The centre is $(-\alpha, -\beta)$ where $\alpha = -g$ and $\beta = -f$.Since the circle c

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(D) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$. The centre is $(-\alpha, -\beta)$ where $\alpha = -g$ and $\beta = -f$.Since the circle cuts $x^2+y^2+2x-3y-5=0$ orthogonally,$2g(1) + 2f(-3/2) = c-5 \implies 2g-3f = c-5$.Since the circle cuts $x^2+y^2-3x+2y+1=0$ orthogonally,$2g(-3/2) + 2f(1) = c+1 \implies -3g+2f = c+1$.Subtracting the two equations: $(2g-3f) - (-3g+2f) = (c-5) - (c+1) \implies 5g-5f = -6 \implies g-f = -6/5$.Since the circle passes through $(1, -1)$,$1+1+2g-2f+c=0 \implies 2g-2f+c = -2$.From $2g-3f = c-5$,we have $c = 2g-3f+5$. Substituting into the circle equation: $2g-2f+(2g-3f+5) = -2 \implies 4g-5f = -7$.We have the system: $g-f = -1.2$ and $4g-5f = -7$. Multiplying the first by $4$: $4g-4f = -4.8$.Subtracting: $(4g-5f) - (4g-4f) = -7 - (-4.8) \implies -f = -2.2 \implies f = 2.2$.Then $g = 2.2 - 1.2 = 1$.Thus,$\alpha = -g = -1$ and $\beta = -f = -2.2$.Therefore,$\alpha-5\beta = -1 - 5(-2.2) = -1 + 11 = 10$.

If $(\alpha, \beta)$ is the centre of the circle which passes through the point $(1, -1)$ and cuts the circles $x^2+y^2+2x-3y-5=0$ and $x^2+y^2-3x+2y+1=0$ orthogonally,then $\alpha-5\beta=$

Similar Questions

The equation of the circle passing through the point $(-2, 4)$ and through the points of intersection of the circle ${x^2} + {y^2} - 2x - 6y + 6 = 0$ and the line $3x + 2y - 5 = 0$ is:

$A$ circle $S = x^2 + y^2 + 2gx + 2fy + 4 = 0$ cuts the circle $x^2 + y^2 - 4x - 4y - 4 = 0$ orthogonally and makes an angle of $60^{\circ}$ with the circle $x^2 + y^2 + 4x + 4y + 4 = 0$. Then the radius of the circle $S = 0$ is

$A$ tangent $PT$ is drawn to the circle $x^2 + y^2 = 4$ at the point $P(\sqrt{3}, 1)$. $A$ line $L$ perpendicular to $PT$ is a tangent to the circle $(x - 3)^2 + y^2 = 1$. The possible equation of $L$ is:

The number of common tangents to the circles $x^2+y^2+4x=0$ and $x^2+y^2-2x=0$ is

The equation of the circle which passes through the point $(3,2)$,bisects the circumference of the circle $x^2+y^2=15$,and cuts the circle $x^2+y^2+4x+6y+3=0$ orthogonally is

Vedclass Test Series

Exam Paper Generator

Online Exam Module