The four distinct points (0, 0), (2, 0), (0, | vedclass

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

Question

(B) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.Since it passes through $(0, 0)$,we get $c = 0$.Since it passes through $(2, 0

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.Since it passes through $(0, 0)$,we get $c = 0$.Since it passes through $(2, 0)$,we get $4 + 4g = 0 \Rightarrow g = -1$.Since it passes through $(0, -2)$,we get $4 - 4f = 0 \Rightarrow f = 1$.The equation of the circle is $x^2 + y^2 - 2x + 2y = 0$.For the point $(k, -2)$ to be on the circle,it must satisfy the equation:$k^2 + (-2)^2 - 2(k) + 2(-2) = 0$$k^2 + 4 - 2k - 4 = 0$$k^2 - 2k = 0$$k(k - 2) = 0$Thus,$k = 0$ or $k = 2$.Since the points must be distinct,$(k, -2)$ cannot be $(0, -2)$,so $k 
eq 0$.Therefore,$k = 2$.

The four distinct points $(0, 0), (2, 0), (0, -2)$ and $(k, -2)$ are concyclic,if $k =$

Similar Questions

The Cartesian equation of the curve given by $x = 6 \cos \theta$ and $y = 6 \sin \theta$ is

Find the equation of a circle with radius $4$ that touches the $x$-axis at a distance of $-3$ from the origin.

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

If the equation $x^{2}+y^{2}-10x+21=0$ has real roots $x=\alpha$ and $y=\beta,$ then

The equation of the circle passing through the origin and cutting intercepts of length $3$ and $4$ units from the positive axes is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module