Question

The least distance of the line 8x-4y+73=0 from the circle 16x^2+16y^2+48x-8y-43=0 is

Answer 1

please see the attachment

Answer 2

The the least distance is $2\sqrt{5}$

Given,

$16x^{2} +16y^{2} +48x-8y-43$= 0

8x-4y+73=0

To Find,

We have to find The least distance of the line 8x-4y+73=0 from the circle $16x^{2} +16y^{2} +48x-8y-43$= 0

Solution,

According to the question,

$16x^{2} +16y^{2} +48x-8y-43$= 0 _____(1)

now, we divided this equation by 16

from that we get,

$x^{2} -y^{2} +3x- \frac{1}{2} y-\frac{43}{16}$=0

so the centre of the circle is - $(\frac{-3}{2} ,\frac{1}{4} )$

Now, we find the radius of the circle

let assume r is radius of the circle,

then

r = $\sqrt{\frac{9}{4} +\frac{1}{16} +\frac{43}{16} }$

= $\sqrt{\frac{(36+1+43)}{16} }$

= $\sqrt{\frac{80}{16} } = \sqrt{5}$

so the radius of the circle is √5.

Now , again according to the question,

8x-4y+73=0

hence,

or, OA= $|\frac{-12-1+73}{\sqrt{80} } |$

=$3\sqrt{5}$

The least distance,

AB = (OA- OB)

= ( $3\sqrt{5}$ - √5 )

= $2\sqrt{5}$

Hence the the least distance of the line 8x-4y+73=0 from the circle $16x^{2} +16y^{2} +48x-8y-43$= 0 is $2\sqrt{5}$

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