A circle is given by x 2 + ( y - 1 ) 2 = 1 , another circle C touches it externally and also the x -axis, then the locus of its centre is [2005]

Question

A circle is given by $x^{2} + {(y - 1)}^{2} = 1$ , another circle $C$ touches it externally and also the $x$ -axis, then the locus of its centre is [2005]

Smart Achievers · Accepted Answer

(4)

Let the centre of circle $C$ be $(h, k)$ . This circle touches $x$ -axis.

$∴$ $radius =$ $| k |$

Also it touches the given circle $x^{2} + {(y - 1)}^{2} = 1$ , with centre (0, 1) and radius 1, externally.

$∴$ Distance between centres = sum of radii

$\Rightarrow \sqrt{{(h - 0)}^{2} + {(k - 1)}^{2}} = 1 +$ $| k |$

$\Rightarrow h^{2} + k^{2} - 2 k + 1 = 1 + 2$ $| k |$ $+ k^{2}$

$\Rightarrow h^{2} = 2 k + 2$ $| k |$

$∴$ $Locus of (h, k) is x^{2} = 2 y + 2$ $| y |$

$Now if y > 0, it becomes x^{2} = 4 y$

$and if y \leq 0, it becomes x = 0$

$∴$ $Combining the two, the required locus is {(x, y) : x^{2} = 4 y} \cup {(0, y) : y \leq 0}$

Solution

Q.
A circle is given by $x^{2} + {(y - 1)}^{2} = 1$ , another circle $C$ touches it externally and also the $x$ -axis, then the locus of its centre is [2005]

1 ${(x, y) : x^{2} = 4 y} \cup {(x, y) : y \leq 0}$

2 ${(x, y) : x^{2} + {(y - 1)}^{2} = 4} \cup {(x, y) : y \leq 0}$

3 ${(x, y) : x^{2} = y} \cup {(0, y) : y \leq 0}$

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

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Solution

Q. A circle is given by x2+(y-1)2=1, another circle C touches it externally and also the x-axis, then the locus of its centre is [2005]

1 {(x,y):x2=4y}∪{(x,y):y≤0}

2 {(x,y):x2+(y-1)2=4}∪{(x,y):y≤0}

3 {(x,y):x2=y}∪{(0,y):y≤0}