The lengths of the intercepts made by a circl | vedclass

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

Question

(B) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$ ... $(i)$ The length of the intercept made by the circle on the $X$-axis is $2\sqrt{g^2-c}

Vedclass · Accepted Answer

$\left(\frac{-4}{3}, \frac{5}{3}\right)$

Vedclass · Answer

(B) Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$ ... $(i)$ The length of the intercept made by the circle on the $X$-axis is $2\sqrt{g^2-c} = \frac{2\sqrt{13}}{3} \Rightarrow g^2-c = \frac{13}{9}$ ... $(ii)$ The length of the intercept made by the circle on the $Y$-axis is $2\sqrt{f^2-c} = \frac{2\sqrt{22}}{3} \Rightarrow f^2-c = \frac{22}{9}$ ... $(iii)$ Given the radius $r = \frac{\sqrt{38}}{3}$,we have $r^2 = g^2+f^2-c = \frac{38}{9}$ ... $(iv)$ From $(ii)$,$g^2 = c + \frac{13}{9}$. From $(iii)$,$f^2 = c + \frac{22}{9}$. Substituting these into $(iv)$: $(c + \frac{13}{9}) + (c + \frac{22}{9}) - c = \frac{38}{9}$ $c + \frac{35}{9} = \frac{38}{9} \Rightarrow c = \frac{3}{9} = \frac{1}{3}$ Now,$g^2 = \frac{1}{3} + \frac{13}{9} = \frac{16}{9} \Rightarrow g = \pm \frac{4}{3}$ And $f^2 = \frac{1}{3} + \frac{22}{9} = \frac{25}{9} \Rightarrow f = \pm \frac{5}{3}$ Since the centre $C(-g, -f)$ lies in the second quadrant,$g$ must be positive and $f$ must be negative. Thus,$C = (-\frac{4}{3}, \frac{5}{3})$.

The lengths of the intercepts made by a circle $S$ on $X$ and $Y$-axes are $\frac{2 \sqrt{13}}{3}$ and $\frac{2 \sqrt{22}}{3}$ respectively. If the radius of the circle $S$ is $\frac{\sqrt{38}}{3}$ and its centre $C$ lies in the second quadrant,then $C=$

Similar Questions

If the power of the point $(1, 6)$ with respect to the circle $x^2 + y^2 + 4x - 6y - a = 0$ is $-16$,then $a =$

If $(1, 1), (-2, 2), (2, -2)$ are $3$ points on a circle $S$,then the perpendicular distance from the centre of the circle $S$ to the line $3x - 4y + 1 = 0$ is

Find the equation of the circle with centre $(-a, -b)$ and radius $\sqrt{a^{2}-b^{2}}$.

The equation of the circle with centre on the $x$-axis,radius $4$ and passing through the origin,is

If $(a, 1/a)$,$(b, 1/b)$,$(c, 1/c)$,and $(d, 1/d)$ are four distinct points on a circle of radius $4$ units,then $abcd$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module