The locus of the centre of the circle which t | vedclass

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

Question

(B) Let the centre of the circle be $(h, k)$ and its radius be $r$. Since the circle touches the $x$-axis,the radius $r = |k|$.Since the circle touche

Vedclass · Accepted Answer

$\{ (x, y) : {x^2} = 4y\} \cup \{ (0, y) : y < 0\} $

Vedclass · Answer

(B) Let the centre of the circle be $(h, k)$ and its radius be $r$. Since the circle touches the $x$-axis,the radius $r = |k|$.Since the circle touches the circle ${x^2} + {(y - 1)^2} = 1$ (centre $(0, 1)$,radius $1$) externally,the distance between their centres is equal to the sum of their radii:$\sqrt {{{(h - 0)}^2} + {{(k - 1)}^2}} = 1 + |k|$Squaring both sides:${h^2} + {(k - 1)^2} = {(1 + |k|)^2}$${h^2} + {k^2} - 2k + 1 = 1 + 2|k| + {k^2}$${h^2} = 2k + 2|k|$Case $1$: If $k > 0$,then $|k| = k$,so ${h^2} = 2k + 2k = 4k$. Thus,the locus is ${x^2} = 4y$ for $y > 0$.Case $2$: If $k Combining these,the locus is $\{ (x, y) : {x^2} = 4y, y > 0\} \cup \{ (0, y) : y < 0\}$.

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

Similar Questions

Let a line perpendicular to the line $2x - y = 10$ touch the parabola $y^2 = 4(x - 9)$ at the point $P$. The distance of the point $P$ from the centre of the circle $x^2 + y^2 - 14x - 8y + 56 = 0$ is ...........

$A$ circle passes through the point $\left( 3, \sqrt{\frac{7}{2}} \right)$ and touches the line pair $x^2 - y^2 - 2x + 1 = 0$. The coordinates of the centre of the circle are:

If the squares of the lengths of the tangents drawn from a point $P$ to the circles $x^{2} + y^{2} = a^2$,$x^2 + y^{2} = b^2$,and $x^{2} + y^{2} = c^{2}$ are in arithmetic progression,then:

The equation of the common tangent to the parabola $y^2=8x$ and the circle $x^2+y^2=2$ is $ax+by+2=0$. If $-\frac{a}{b} > 0$,then $3a^2+2b+1=$

Let $C$ be the largest circle centred at $(2,0)$ and inscribed in the ellipse $\frac{x^2}{36}+\frac{y^2}{16}=1$. If $(1, \alpha)$ lies on $C$,then $10 \alpha^2$ is equal to $.........$

Vedclass Test Series

Exam Paper Generator

Online Exam Module