Question

find the radius of the circle (x cos alpha+y sin alpha -a)^2+(x sin alpha-y cos alpha-b)^2=k^2 and if alpha varies ,find the locus of its centre.

Answer 1

Given:

Equation of circle,$( xcos\alpha +ysin\alpha-a)^{2} +(xsin\alpha -ycos\alpha -b)^{2}=k^{2}$

To find:

We have to find the radius of the circle and locus of its centre

Solution:

Equation of circle,

$(x^{2}cos^{2}\alpha +y^{2}sin^{2}\alpha +a^{2}+2xysin\alpha cos\alpha -2aysin\alpha -2axcos\alpha ) +(x^{2}sin^{2}\alpha +y^{2}cos^{2}\alpha +b^{2}-2xysin\alpha cos\alpha +2bycos\alpha -2bxsin\alpha )=k^{2}$

$x^{2} (sin^{2}\alpha +cos^{2} \alpha)+y^{2}(sin^{2}\alpha +cos^{2} \alpha)+x(-2acos\alpha -2bsin\alpha )+y(-2asin\alpha +2bcos\alpha )+a^{2}+b^{2}=k^{2}$

$x^{2} +y^{2}+2x(-acos\alpha -bsin\alpha )+2y(-asin\alpha +bcos\alpha )+a^{2} +b^{2} -k^{2} =0$

comparing above equation with general equation of circle $x^{2} +y^{2}+2gx+2fy+c=0$  we get,

$g=-acos\alpha -bsin\alpha$

$f=-asin\alpha +bcos\alpha$

$c=a^{2} +b^{2} -k^{2}$

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

                                = $\sqrt{(-acos\alpha -bsin\alpha )^{2}+(-asin\alpha +bcos\alpha )^{2} -a^{2} -b^{2} +k^{2} }$

                                = $\sqrt{k^{2} }$

                                = $k$

∴The radius of the given circle = k

Locus of centre,

$\sqrt{g^{2} +f^{2}-c }$ =  $k$

On squaring both sides we get,

$g^{2} +f^{2} -c=k^{2}$

$g^{2} +f^{2} -(a^{2} +b^{2} -k^{2} )=k^{2}$

$g^{2} +f^{2} -a^{2} -b^{2} +k^{2} =k^{2}$

$g^{2} +f^{2}= a^{2} +b^{2}$ This is the required locus centre of given circle

where centre= $(-g,-f)$

Answer 2

Answer:

Mark as the brainless

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