If the equation of the circle whose radius is | vedclass

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

Question

(A) Let the required circle be $S_2 = 0$. The point of contact is $P(1,2)$. The center of the given circle $S_1: x^2+y^2+2x+8y-23=0$ is $C_1(-1,-4)$ a

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Let the required circle be $S_2 = 0$. The point of contact is $P(1,2)$. The center of the given circle $S_1: x^2+y^2+2x+8y-23=0$ is $C_1(-1,-4)$ and its radius $r_1 = \sqrt{(-1)^2+(-4)^2-(-23)} = \sqrt{1+16+23} = \sqrt{40} = 2\sqrt{10}$.Since the circles touch externally,the distance between centers $C_1C_2 = r_1 + r_2 = 2\sqrt{10} + \sqrt{10} = 3\sqrt{10}$.The center $C_2(h,k)$ lies on the line joining $C_1(-1,-4)$ and $P(1,2)$. The vector $\vec{C_1P} = (1-(-1), 2-(-4)) = (2,6)$.The unit vector along $\vec{C_1P}$ is $\frac{(2,6)}{\sqrt{2^2+6^2}} = \frac{(2,6)}{\sqrt{40}} = \frac{(2,6)}{2\sqrt{10}} = (\frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}})$.Since $C_2$ is at a distance $r_2 = \sqrt{10}$ from $P(1,2)$ along the direction $\vec{C_1P}$,we have $C_2 = P + r_2 	imes (	ext{unit vector}) = (1,2) + \sqrt{10}(\frac{1}{\sqrt{10}}, \frac{3}{\sqrt{10}}) = (1+1, 2+3) = (2,5)$.The equation of the circle with center $(2,5)$ and radius $\sqrt{10}$ is $(x-2)^2 + (y-5)^2 = (\sqrt{10})^2$.$x^2-4x+4 + y^2-10y+25 = 10$.$x^2+y^2-4x-10y+19 = 0$.Comparing with $x^2+y^2+ax+by+c=0$,we get $a=-4, b=-10, c=19$.$|a+b+c| = |-4-10+19| = |5| = 5$.

If the equation of the circle whose radius is $\sqrt{10}$ and which touches the circle $x^2+y^2+2x+8y-23=0$ externally at the point $(1,2)$ is $x^2+y^2+ax+by+c=0$,then $|a+b+c|=$

Similar Questions

$x^2+y^2+2x-6y-6=0$ and $x^2+y^2-6x-2y+k=0$ are two intersecting circles and $k$ is not an integer. If $\theta$ is the angle between the two circles and $\cos \theta = \frac{-5}{24}$,then $k=$

If the lengths of the chords intercepted by the circle $x^2 + y^2 + 2gx + 2fy = 0$ from the coordinate axes are $10$ and $24$ respectively,then the radius of the circle is:

If the circle $(x - h)^2 + (y - k)^2 = r^2$ touches the curve $y = x^2 + 1$ at the point $(1, 2)$,then the possible locations of the points $(h, k)$ are given by

Position of the point $(1, 1)$ with respect to the circle $x^2 + y^2 - x + y - 1 = 0$ is

The number of common tangents to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 6x - 8y = 24$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module