Question

the least distance of the line 8x-4y+73=0 from the circle 16xsqare +16ysquare+48x-8y-43=0is​

Answer 1

The equation of circle given is,

$\longrightarrow 16x^2+16y^2+48x-8y-43=0$

Dividing by 16,

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

$\longrightarrow x^2+3x+\dfrac{9}{4}+y^2-\dfrac{1}{2}\,y+\dfrac{1}{16}=\dfrac{9}{4}+\dfrac{1}{16}+ \dfrac{43}{16}$

$\longrightarrow \left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{4}\right)^2=5$

This implies center of circle is at $\left(- \dfrac{3}{2},\ \dfrac{1}{4}\right)$ and radius of circle is $\sqrt5 .$

Now we need to find perpendicular distance of this center from the given line, from which when the radius of the circle is subtracted, we get the least distance of the line from the circle. (See fig.)

The perpendicular distance of the center of the circle from the given line,

$\longrightarrow p=\dfrac{\left|8\left(-\dfrac{3}{2}\right)-4\left(\dfrac{1}{4}\right)+73\right|}{\sqrt{8^2+(-4)^2}}$

$\longrightarrow p=3\sqrt5$

Now the least distance will be,

$\longrightarrow d=p-\sqrt5$

$\longrightarrow\underline{\underline{d =2\sqrt5}}$

Answer 2

Concept Used :-

How to find Centre and radius of circle

$\rm \: If \: the \: equation \: of \: circle \: is \: given \: by \:$

$\bigstar \: \: \boxed{ \pink{\rm \: {x}^{2} + {y}^{2} + 2gx + 2fy + c = 0 \: then \: }}$

$\rm \: Centre \: is \: \: ( - \dfrac{1}{2}coeff. \: of \: x , - \dfrac{1}{2}coeff. \: of \: y )$

$\bigstar \: \: \boxed{ \pink{ \therefore \: \rm \: Centre \: is \: ( - g, - f)}}$

and

$\bigstar \: \: \boxed{ \pink{\rm \: Radius \: is \: \sqrt{ {g}^{2} + {f}^{2} - c } }}$

Distance between point and line:-

Let us consider a line ax + by + c = 0 and let P(p, q) be any point in plane, then shortest distance between line and point is perpendicular distance and is given by

$\bigstar \: \: \boxed{ \pink{ \rm \: perpendicular \: distance \: (d) = \dfrac{ |ap + bq + c| }{ \sqrt{ {a}^{2} + {b}^{2} } } }}$

Note :-

• Before finding the centre and radius of circle, the coefficients of x^2 and y^2 must be unity.

$\large\underline\purple{\bold{Solution :- }}$

Given

• Equation of circle is

$\rm : \implies \: {16x}^{2} + {16y}^{2} + 48x - 8y - 43 = 0$

Divide by 16, we get

$\rm : \implies \: {x}^{2} + {y}^{2} + 3x - \dfrac{1}{2} y - \dfrac{43}{8} = 0$

So,

• Centre of circle is evaluated by using formula,

$\bigstar \: \: \boxed{ \pink{ \rm : \implies \:Centre \: is \: ( - \: \dfrac{3}{2} , \dfrac{1}{4} )}}$

and

• Radius of circle is given by

$\rm : \implies \:r \: = \sqrt{ {( - \dfrac{3}{2} )}^{2} + {(\dfrac{1}{4}) }^{2} +\dfrac{43}{16} }$

$\rm : \implies \:r \: = \sqrt{\dfrac{9}{4} + \dfrac{1}{16} + \dfrac{43}{16} }$

$\rm : \implies \:r \: = \sqrt{\dfrac{36 + 1 + 43}{16} }$

$\rm : \implies \:r \: = \sqrt{\dfrac{80}{16} }$

$\bigstar \: \: \boxed{ \pink{ \rm : \implies \:r \: = \: \sqrt{5} }}$

Now,

Equation of line is

• 8x - 4y + 73 = 0.

So, perpendicular distance between line and centre is given by

$\rm : \implies \:d \: = \: \dfrac{ |8 \times (\dfrac{ - 3}{2}) - 4 \times \dfrac{1}{4} + 73| }{ \sqrt{ {8}^{2} + {( - 4)}^{2} } }$

$\rm : \implies \:d \: = \: \dfrac{ | - 12 - 1 + 73| }{ \sqrt{64 + 16} }$

$\rm : \implies \:d \: = \dfrac{60}{ \sqrt{80} }$

$\rm : \implies \:d \: = \dfrac{60}{4 \sqrt{5} }$

$\rm : \implies \:d \: = \dfrac{15}{ \sqrt{5} } \times \dfrac{ \sqrt{5} }{ \sqrt{5} }$

$\bigstar \: \: \boxed{ \pink{\rm : \implies \:d \: = \: 3 \sqrt{5} \: units}}$

So,

Now the least distance between circle and line is given by

$\rm : \implies \:least \: distance \: = \: d \: - \: r$

$\rm : \implies \:least \: distance \: = \: 3 \sqrt{5} \: - \: \sqrt{5}$

$\bigstar \: \: \boxed{ \pink{ \rm : \implies \:least \: distance \: = 2 \sqrt{5} \: units}}$