Question

Consider the relation $4{{l}^{2}}-5{{m}^{2}}+6l+1=0$, where l,m belongs to a set of real numbers . The number of tangents which can be drawn from the point ( 2, -3 ) to the above fixed circle are
( a ) 0
( b ) 1
( c ) 2
( d ) 1 or 2

Answer 1

Hint: Firstly, we will find the values of f, g and c by comparing relations we get from $\dfrac{\left| -gl=mf+1 \right|}{\sqrt{{{l}^{2}}+{{m}^{2}}}}=\sqrt{{{g}^{2}}+{{f}^{2}}-c}$ and $4{{l}^{2}}-5{{m}^{2}}+6l+1=0$. Then we will find the equation of the circle by putting values of f and g and c. Then, we will evaluate distance between points ( 2, -3 ) and ( 3, 0 ) and then we will decide the number of tangents which can be drawn on the circle from fixed points.

Complete step-by-step answer:
Let consider the general equation of the circle be ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ ….. ( i )
So, we know that if we have line lx + my +1 = 0 will touch circle if the length of perpendicular from the centre ( -g, -f ) of the circle on the line is equal to its radius that is,
\[\dfrac{\left| -gl=mf+1 \right|}{\sqrt{{{l}^{2}}+{{m}^{2}}}}=\sqrt{{{g}^{2}}+{{f}^{2}}-c}\]
Or, using cross multiplication we get
$\left| -gl=mf+1 \right|=\sqrt{{{g}^{2}}+{{f}^{2}}-c}\times \sqrt{{{l}^{2}}+{{m}^{2}}}$
Squaring both sides we get
${{\left( -gl=mf+1 \right)}^{2}}=\left( {{g}^{2}}+{{f}^{2}}-c \right)\times \left( {{l}^{2}}+{{m}^{2}} \right)$
Solving brackets on both sides we get
${{g}^{2}}{{l}^{2}}+{{m}^{2}}{{f}^{2}}+1-2glmf+2mf-2gl={{g}^{2}}{{l}^{2}}+{{g}^{2}}{{m}^{2}}+{{f}^{2}}{{l}^{2}}+{{f}^{2}}{{m}^{2}}-c{{l}^{2}}-c{{m}^{2}}$
Taking common terms out from the equation, we get
\[(c-{{f}^{2}}){{l}^{2}}+(c-{{g}^{2}}){{m}^{2}}-2gl-2fm+2fglm+1=0\] ….( ii )
In question we have relation, $4{{l}^{2}}-5{{m}^{2}}+6l+1=0$…..( iii )
Comparing equation ( ii ) and ( iii ), we get
\[(c-{{f}^{2}})=4,(c-{{g}^{2}})=-5,-2g=6,-2f=0,2fg=0,\]
Solving further, we get -2f = 0, -2g = 6 and ( c – 0 ) = 4
So, f = 0, g = -3 and c = 4
Substituting values in equation ( i ), we get
Equation of circle as, ${{x}^{2}}+{{y}^{2}}-6x+4=0$ whose centre is ( 3, 0 ) and radius will be \[\sqrt{{{(-3)}^{2}}+{{(0)}^{2}}-(4)}\]that is radius is \[\sqrt{5}\] units.
Now, we know that distance formula between two points is $\sqrt{{{(c-a)}^{2}}+{{(d-b)}^{2}}}$ , where points are ( a, b ) and ( c, d )
So, distance between point ( 2, -3 ) and ( 3, 0) will be $\sqrt{{{(3-2)}^{2}}+{{(0-(-3))}^{2}}}$ which is $\sqrt{10}$ units,
Now, as $\sqrt{10}>\sqrt{5}$, so ( 2,-3 ) lies outside the circle and we know we can draw maximum 2 tangents from a point which lies outside the circle.

So, the correct answer is “Option C”.

Note: To solve such questions, always remember that general equation of circle is of form ${{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0$ also, if we have line lx + my +1 = 0 will touch circle if the length of perpendicular from the centre ( -g, -f ) of the circle on the line is equal to its radius that is, $\dfrac{\left| -gl=mf+1 \right|}{\sqrt{{{l}^{2}}+{{m}^{2}}}}=\sqrt{{{g}^{2}}+{{f}^{2}}-c}$. And remember that we can draw maximum 2 tangents from a point which lies outside the circle.