The number of circles touching the lines x = | vedclass

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

Question

(B) Let the circle have center $(h, k)$ and radius $r$.Since the circle touches the lines $y = a$ and $y = b$,the diameter of the circle must be equal

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Let the circle have center $(h, k)$ and radius $r$.Since the circle touches the lines $y = a$ and $y = b$,the diameter of the circle must be equal to the distance between these two parallel lines.Thus,$2r = |b - a|$,which implies $r = \frac{|b - a|}{2}$.The distance from the center $(h, k)$ to the line $y = a$ is $|k - a| = r$,so $k = a \pm r$.The distance from the center $(h, k)$ to the line $x = 0$ is $|h| = r$,so $h = \pm r$.Since $r$ is fixed,for each value of $h$ ($r$ or $-r$),we get a circle. Thus,there are exactly $2$ such circles.

The number of circles touching the lines $x = 0$,$y = a$ and $y = b$ is

Similar Questions

The centre of the circle $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$ is

Find the centre and radius of the circle $(x+5)^{2}+(y-3)^{2}=36$.

$A$ circle having its centre at the origin passes through the three vertices of an equilateral triangle,the length of whose median is $9$ units. Then,the equation of that circle is

The parametric equations of the circle $x^2+y^2+2x-4y-4=0$ are

If $(1, 1), (-2, 2), (2, -2)$ are $3$ points on a circle $S$,then the perpendicular distance from the centre of the circle $S$ to the line $3x - 4y + 1 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module