Circles are drawn through the point (2, 0) to | 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 $x^2 + y^2 + 2gx + 2fy + c = 0$.Since it passes through $(2, 0)$,we have $4 + 4g + c = 0$,which implies $c = -4g

Vedclass · Accepted Answer

$x^2 + y^2 - 9x + 2fy + 14 = 0$

Vedclass · Answer

(C) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since it passes through $(2, 0)$,we have $4 + 4g + c = 0$,which implies $c = -4g - 4$.The length of the intercept on the $x$-axis is given by $2\sqrt{g^2 - c} = 5$.Squaring both sides,$4(g^2 - c) = 25$.Substituting $c = -4g - 4$,we get $4(g^2 + 4g + 4) = 25$.$4g^2 + 16g + 16 = 25 \Rightarrow 4g^2 + 16g - 9 = 0$.Solving for $g$,$(2g + 9)(2g - 1) = 0$,so $g = -\frac{9}{2}$ or $g = \frac{1}{2}$.The centre of the circle is $(-g, -f)$. Since the centre lies in the first quadrant,$-g > 0$,so $g Thus,we choose $g = -\frac{9}{2}$.Then $c = -4(-\frac{9}{2}) - 4 = 18 - 4 = 14$.The equation becomes $x^2 + y^2 - 9x + 2fy + 14 = 0$.

Circles are drawn through the point $(2, 0)$ to cut an intercept of length $5$ units on the $x$-axis. If their centres lie in the first quadrant,then their equation is

Similar Questions

Find the equation of the circle which passes through the point $(-2, -7)$ and is concentric with the circle $x^{2} + y^{2} - 8x + 6y - 5 = 0$.

In a square $ABCD$ of side length $a$,suppose $AB$ and $AD$ are along the coordinate axes. Then,the circle that circumscribes the square is

The circle possessing $y$-axis as its tangent at $(0,2)$ and passing through $(-1,0)$,also passes through

Find the equation of the circle passing through the point $(3, 6)$ and having its center at $(2, -1)$.

Find the centre and radius of the circle $x^{2}+y^{2}-8x+10y-12=0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module