If the circle x^{2}+y^{2}-2gx+6y-19c=0,where | vedclass

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

Question

(D) The equation of the circle is $x^{2}+y^{2}-2gx+6y-19c=0$. Since the circle passes through $(6,1)$,we have: $6^{2}+1^{2}-2g(6)+6(1)-19c=0$ $36+1-12

Vedclass · Accepted Answer

$2\sqrt{23}$

Vedclass · Answer

(D) The equation of the circle is $x^{2}+y^{2}-2gx+6y-19c=0$. Since the circle passes through $(6,1)$,we have: $6^{2}+1^{2}-2g(6)+6(1)-19c=0$ $36+1-12g+6-19c=0$ $43-12g-19c=0 \implies 12g+19c=43$ $(1)$ The centre of the circle is $(g, -3)$. Since the centre lies on the line $x-2cy=8$,we have: $g-2c(-3)=8 \implies g+6c=8$ $(2)$ From $(2)$,$g=8-6c$. Substituting into $(1)$: $12(8-6c)+19c=43$ $96-72c+19c=43$ $-53c = -53 \implies c=1$ Substituting $c=1$ into $g=8-6c$,we get $g=8-6=2$. The equation of the circle is $x^{2}+y^{2}-4x+6y-19=0$. The length of the intercept on the $x$-axis is given by $2\sqrt{g^{2}-c'}$ where $c'$ is the constant term $-19$. Length $= 2\sqrt{g^{2}-(-19)} = 2\sqrt{2^{2}+19} = 2\sqrt{4+19} = 2\sqrt{23}$.

If the circle $x^{2}+y^{2}-2gx+6y-19c=0$,where $g, c \in R$,passes through the point $(6,1)$ and its centre lies on the line $x-2cy=8$,then the length of the intercept made by the circle on the $x$-axis is:

Similar Questions

$A$ circle cuts off positive intercepts $5$ and $6$ on the $x$ and $y$ axes respectively,and passes through the origin. Then the equation of the circle is

Find the equation of the circle passing through the foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ and having its center at $(0, 3)$.

Find the equation of the circle with centre $(-3, 2)$ and radius $4$.

The centre of a circle is $(2, -3)$ and the circumference is $10 \pi$. Then its equation is

Find the radius and center of the circle $2x^2 + 2y^2 = 3x - 5y + 7$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module