(1, k) is a point on the circle passing throu | 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 $x^2 + y^2 + 2gx + 2fy + c = 0$. Since the points $(-1, 1), (0, -1)$,and $(1, 0)$ lie on the circle: For $(-1, 1

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(B) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. Since the points $(-1, 1), (0, -1)$,and $(1, 0)$ lie on the circle: For $(-1, 1): 1 + 1 - 2g + 2f + c = 0 \Rightarrow -2g + 2f + c = -2 \dots (i)$ For $(0, -1): 0 + 1 + 0 - 2f + c = 0 \Rightarrow -2f + c = -1 \dots (ii)$ For $(1, 0): 1 + 0 + 2g + 0 + c = 0 \Rightarrow 2g + c = -1 \dots (iii)$ Adding $(ii)$ and $(iii)$,we get $2c = -2 \Rightarrow c = -1$. Substituting $c = -1$ into $(iii)$,$2g - 1 = -1 \Rightarrow g = 0$. Substituting $c = -1$ into $(ii)$,$-2f - 1 = -1 \Rightarrow f = 0$. The equation of the circle is $x^2 + y^2 - 1 = 0$. Since $(1, k)$ lies on the circle: $1^2 + k^2 - 1 = 0 $ $\Rightarrow k^2 = 0 $ $\Rightarrow k = 0$. However,the problem states $k 
eq 0$. Re-evaluating the points: For $(-1, 1): 1 + 1 - 2g + 2f + c = 0 \Rightarrow -2g + 2f + c = -2$. For $(0, -1): 1 - 2f + c = 0 \Rightarrow -2f + c = -1$. For $(1, 0): 1 + 2g + c = 0 \Rightarrow 2g + c = -1$. Adding $(ii)$ and $(iii)$: $2c = -2 \Rightarrow c = -1$. $2g = -1 - (-1) = 0 \Rightarrow g = 0$. $-2f = -1 - (-1) = 0 \Rightarrow f = 0$. The circle is $x^2 + y^2 = 1$. The point $(1, k)$ on this circle implies $1 + k^2 = 1 \Rightarrow k = 0$. Given the options,there is a discrepancy in the provided points or the question statement. Assuming the intended circle equation is $x^2 + y^2 - x - y - 1 = 0$ or similar,but based on the provided points,$k=0$ is the only solution.

$(1, k)$ is a point on the circle passing through the points $(-1, 1), (0, -1)$ and $(1, 0)$. If $k \neq 0$,then $k =$

Similar Questions

Let the abscissae of the two points $P$ and $Q$ be the roots of $2x^{2}-rx+p=0$ and the ordinates of $P$ and $Q$ be the roots of $y^{2}-sy-q=0$. If the equation of the circle described on $PQ$ as diameter is $2(x^{2}+y^{2})-11x-14y-22=0$,then $2r+s-2q+p$ is equal to

The Cartesian equation of the curve $x=3+5 \cos \theta, y=2+5 \sin \theta$ is $(0 \leq \theta \leq 2 \pi)$.

The equations of the circles touching both the axes and passing through the point $(1, 2)$ are

If the lines $2x + 3y + 1 = 0$ and $3x - y - 4 = 0$ lie along diameters of a circle of circumference $10\pi$,then the equation of the circle is

The number of integral values of $K$ for which $x^2+y^2+kx+(1-k)y+5=0$ represents a circle whose radius cannot exceed $5$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module