If the inverse point of the point (3, 2) with | vedclass

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

Question

(C) The given circle is $x^2+y^2-2x+4y-4=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1, f=2, c=-4$. The center $(h, k)$ is $(-g, -f) = (1, -2)

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) The given circle is $x^2+y^2-2x+4y-4=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $g=-1, f=2, c=-4$. The center $(h, k)$ is $(-g, -f) = (1, -2)$ and the radius $r = \sqrt{g^2+f^2-c} = \sqrt{1+4+4} = 3$. The inverse point $(l, m)$ of a point $P(\alpha, \beta)$ with respect to a circle with center $(h, k)$ and radius $r$ is given by: $l = h + \frac{r^2(\alpha-h)}{(\alpha-h)^2+(\beta-k)^2}$ and $m = k + \frac{r^2(\beta-k)}{(\alpha-h)^2+(\beta-k)^2}$. Here,$(\alpha, \beta) = (3, 2)$,$(h, k) = (1, -2)$,and $r^2 = 9$. Calculate the denominator: $(\alpha-h)^2 + (\beta-k)^2 = (3-1)^2 + (2+2)^2 = 2^2 + 4^2 = 4 + 16 = 20$. Thus,$\lambda = \frac{r^2}{(\alpha-h)^2+(\beta-k)^2} = \frac{9}{20}$. $l = 1 + \frac{9}{20}(3-1) = 1 + \frac{18}{20} = 1 + 0.9 = 1.9 = \frac{38}{20}$. $m = -2 + \frac{9}{20}(2+2) = -2 + \frac{36}{20} = -2 + 1.8 = -0.2 = -\frac{4}{20}$. Now,calculate $2l + 19m = 2(\frac{38}{20}) + 19(-\frac{4}{20}) = \frac{76}{20} - \frac{76}{20} = 0$.

If the inverse point of the point $(3, 2)$ with respect to the circle $x^2+y^2-2x+4y-4=0$ is $(l, m)$,then $(2l+19m) =$

Similar Questions

The circle $x^2 + y^2 - 4x - 4y + 4 = 0$ is...

If the circles $x^2 + y^2 - 9 = 0$ and $x^2 + y^2 + 2ax + 2y + 1 = 0$ touch each other,then $a =$

The point on the circle $x^2+y^2=4$ whose distance from the line $4x+3y-12=0$ is $4/5$ units is equal to

If a circle with center $(-1, 1)$ touches the line $x + 2y + 4 = 0$, then the coordinates of the point of contact are

If the angle between the circles $x^2+y^2-12x-6y+41=0$ and $x^2+y^2+kx+6y-59=0$ is $45^{\circ}$,then a value of $k$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module