The locus of the centre of the circle which touches the circle ${x^2} + {(y - 1)^2} = 1$ externally and also touches the $x$-axis is

  • A
    $\{ (x, y) : {x^2} + {(y - 1)^2} = 4\} \cup \{ (x, y) : y < 0\} $
  • B
    $\{ (x, y) : {x^2} = 4y\} \cup \{ (0, y) : y < 0\} $
  • C
    $\{ (x, y) : {x^2} = y\} \cup \{ (0, y) : y < 0\} $
  • D
    $\{ (x, y) : {x^2} = 4y\} \cup \{ (x, y) : y < 0\} $

Explore More

Similar Questions

The circle ${x^2} + {y^2} = 4$ cuts the line joining the points $A(1, 0)$ and $B(3, 4)$ in two points $P$ and $Q$. Let $\frac{BP}{PA} = \alpha$ and $\frac{BQ}{QA} = \beta$. Then $\alpha$ and $\beta$ are roots of the quadratic equation

Difficult
View Solution

The power of the point $B(-1, 1)$ with respect to the circle $S \equiv x^2+y^2-2x-4y+3=0$ is $p$. If the length of the tangent drawn from $B$ to the circle $S=0$ is $t$,then the point $(2, 3)$ with respect to the circle $S^{\prime}=0$ having centre at $(p, t^2)$ and passing through the origin:

$B$ and $C$ are fixed points having coordinates $(3, 0)$ and $(-3, 0)$ respectively. If the vertical angle $\angle BAC$ is $90^o$,then the locus of the centroid of the $\Delta ABC$ has the equation:

$A$ variable circle passes through the fixed point $A(p, q)$ and touches the $x$-axis. The locus of the other end of the diameter through $A$ is

If two circles of the same radius $a$ and centers at $(2, 3)$ and $(5, 6)$ are orthogonal,find the value of $a$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo