Let the centre of a circle,passing through th | vedclass

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

Question

(D) Let the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$. Since it passes through $(0,0)$ and $(1,0)$,the radius $r$ is the distance from $(h,k

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(D) Let the equation of the circle be $(x-h)^2 + (y-k)^2 = r^2$. Since it passes through $(0,0)$ and $(1,0)$,the radius $r$ is the distance from $(h,k)$ to $(0,0)$,so $r^2 = h^2+k^2$.Expanding the equation: $x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = h^2+k^2$,which simplifies to $x^2+y^2-2hx-2ky=0$.Since the circle passes through $(1,0)$,we substitute $x=1, y=0$: $1^2+0^2-2h(1)-2k(0)=0$,which gives $1-2h=0$,so $h=1/2$.The circle touches $x^2+y^2=9$ (a circle with center $(0,0)$ and radius $3$). The distance between the centers is equal to the difference of the radii (since the smaller circle is inside the larger one): $\sqrt{h^2+k^2} = 3 - r = 3 - \sqrt{h^2+k^2}$.Thus,$2\sqrt{h^2+k^2} = 3$,so $\sqrt{h^2+k^2} = 3/2$.Squaring both sides,$h^2+k^2 = 9/4$.Therefore,$4(h^2+k^2) = 4(9/4) = 9$.

Let the centre of a circle,passing through the points $(0,0)$ and $(1,0)$ and touching the circle $x^2+y^2=9$,be $(h, k)$. Then for all possible values of the coordinates of the centre $(h, k)$,$4(h^2+k^2)$ is equal to .............

Similar Questions

The number of common tangents that can be drawn to the circles $x^2+y^2-2x-2y-23=0$ and $x^2+y^2-4x-4y-1=0$ is

Consider the family of circles $x^2+y^2-2x-2\lambda y-8=0$. This family passes through two fixed points $A$ and $B$. Find the distance between these two points.

The condition for the coaxial system $x^2+y^2+2 \lambda x+c=0$,where $\lambda$ is a parameter and $c$ is a constant,to have distinct limiting points,is

If the circle $x^2+y^2+2 \alpha x+c=0$ lies completely inside the circle $x^2+y^2+2 \beta x+c=0$,then which of the following holds?

The point $(3, -4)$ lies on both the circles $x^2 + y^2 - 2x + 8y + 13 = 0$ and $x^2 + y^2 - 4x + 6y + 11 = 0$. Then,the angle between the circles is

Vedclass Test Series

Exam Paper Generator

Online Exam Module