The length of the chord intercepted by the ci | vedclass

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

Question

(D) The equation of the circle is $x^2+y^2-4x+4y+3=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-2, f=2, c=3$. The center of the circle is $C(-g

Vedclass · Accepted Answer

$2\sqrt{10}$

Vedclass · Answer

(D) The equation of the circle is $x^2+y^2-4x+4y+3=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-2, f=2, c=3$. The center of the circle is $C(-g, -f) = (2, -2)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{(-2)^2+2^2-3} = \sqrt{4+4-3} = \sqrt{5}$. The line equation is $x-3y-13=0$. The perpendicular distance $d$ from the center $(2, -2)$ to the line $x-3y-13=0$ is $d = \frac{|(1)(2) - 3(-2) - 13|}{\sqrt{1^2+(-3)^2}} = \frac{|2+6-13|}{\sqrt{10}} = \frac{|-5|}{\sqrt{10}} = \frac{5}{\sqrt{10}} = \frac{\sqrt{10}}{2}$. The length of the chord is $2\sqrt{r^2-d^2} = 2\sqrt{5 - \frac{10}{4}} = 2\sqrt{5 - 2.5} = 2\sqrt{2.5} = 2\sqrt{\frac{5}{2}} = 2 \cdot \frac{\sqrt{5}}{\sqrt{2}} = \sqrt{2} \cdot \sqrt{5} = \sqrt{10}$.

The length of the chord intercepted by the circle $x^2+y^2-4x+4y+3=0$ on the line $x=3y+13$ is units.

Similar Questions

If the length of the chord of the circle $x^{2}+y^{2}=r^{2}$ $(r>0)$ along the line $y-2x=3$ is $r$,then $r^{2}$ is equal to

If the power of a point $(4, 2)$ with respect to the circle $x^2 + y^2 - 2\alpha x + 6y + \alpha^2 - 16 = 0$ is $9$,then the sum of the lengths of all possible intercepts made by such circles on the coordinate axes is:

If the line $x-2y=m$ $(m \in \mathbb{Z})$ intersects the circle $x^2+y^2=2x+4y$ at two distinct points,then the number of possible values of $m$ are

Given three collinear points $A(3,1)$,$B(7,-1)$,and $C(5,0)$. The length of a tangent drawn from $A$ to any circle that passes through $B$ and $C$ is ....... units.

The area of the curve $x^2 + y^2 = 2ax$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module