A point which lies on the circle passing thro | vedclass

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

Question

(C) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. Substituting the points $(1,1), (-6,0),$ and $(-2,2)$ into the equation: $1$) F

Vedclass · Accepted Answer

$(-2,-8)$

Vedclass · Answer

(C) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. Substituting the points $(1,1), (-6,0),$ and $(-2,2)$ into the equation: $1$) For $(1,1): 1 + 1 + 2g + 2f + c = 0 \implies 2g + 2f + c = -2$. $2$) For $(-6,0): 36 + 0 - 12g + 0 + c = 0 \implies -12g + c = -36$. $3$) For $(-2,2): 4 + 4 - 4g + 4f + c = 0 \implies -4g + 4f + c = -8$. Subtracting $(1)$ from $(3)$: $(-4g - 2g) + (4f - 2f) + (c - c) = -8 - (-2) \implies -6g + 2f = -6 \implies -3g + f = -3 \implies f = 3g - 3$. From $(2)$,$c = 12g - 36$. Substitute $f$ and $c$ into $(1)$: $2g + 2(3g - 3) + (12g - 36) = -2 \implies 2g + 6g - 6 + 12g - 36 = -2 \implies 20g = 40 \implies g = 2$. Then $f = 3(2) - 3 = 3$ and $c = 12(2) - 36 = -12$. The equation of the circle is $x^2 + y^2 + 4x + 6y - 12 = 0$. Testing option $(C) (-2,-8)$: $(-2)^2 + (-8)^2 + 4(-2) + 6(-8) - 12 = 4 + 64 - 8 - 48 - 12 = 68 - 68 = 0$. Thus,the point $(-2,-8)$ lies on the circle.

$A$ point which lies on the circle passing through the points $(1,1), (-6,0),$ and $(-2,2)$ is

Similar Questions

The equation obtained by transforming $x^2+y^2-6x+10y-2=0$ to the parallel axes through $(3,-5)$ is

The equation of the circle concentric with the circle $x^2+y^2-6x+12y+15=0$ and having double its area is:

The equation of the circumcircle of the triangle formed by the lines $x = 0$,$y = 0$,and $2x + 3y = 5$ is

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

Find the equation of the circle whose center is $(3, 5)$ and radius is $4$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module