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The least distance of the line 8x -4y+73...

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

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The correct Answer is:
`2sqrt(5)`

`16x^(2)+16y^(2)+48x-8y-43=0`
or `x^(2)+y^(2)+3x-(1)/(2)y-(43)/(16)=0`
Center of the circle `: (-(3)/(2),(1)/(4))`
Radius`= sqrt((9)/(4)+(1)/(16)+(43)/(16))=sqrt(5)`
Required least distance
`=` Distance of line from centre of circle `-` Radius of circle
`= (|8(-(3)/(2))-4((1)/(4))+73|)/(sqrt(64+16))-sqrt(5)`

`=(60)/(4 sqrt(5))-sqrt(5)= 2 sqrt(5)`
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