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Consider a family of circles which are passing through the point (–1, 1) and are tangent to $x-$ axis. If $(h,k)$ are the coordinate of the centre of the circles, then the set of values of $k$ is given by the interval

Option 1)
$-\frac{1}{2}\leq k\leq \frac{1}{2}\;$

Option 2)
$\; k\leq \frac{1}{2}\;$

Option 3)
$\; 0\leq k\leq \frac{1}{2}\;$

Option 4)
$\; k\geq \frac{1}{2}$

Answers (1)

best_answer

As we learnt in

General form of a circle -

$x^{2}+y^{2}+2gx+2fy+c= 0$

- wherein

centre = $\left ( -g,-f \right )$

radius = $\sqrt{g^{2}+f^{2}-c}$

Equation of circle with center $(h,k)$ is:

$(x-h)^{2}+(y-k)^{2}=h^{2}$

Radius of circle is $k$

$(-1-h)^{2}+(1-k)^{2}=k^{2}$

$1^{2}+h^{2}+2h+1+1+k^{2}-2k=k^{2}$

$h^{2}+2h-2k+2=0$

$\Delta \geqslant 0$

We get, $4-4(-2k+2) \geqslant 0$

$k \geqslant \frac{1}{2}$

Option 1)

$-\frac{1}{2}\leq k\leq \frac{1}{2}\;$

This option is incorrect.

Option 2)

$\; k\leq \frac{1}{2}\;$

This option is incorrect.

Option 3)

$\; 0\leq k\leq \frac{1}{2}\;$

This option is incorrect.

Option 4)

$\; k\geq \frac{1}{2}$

This option is correct.

Posted by

divya.saini

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