# If $3{\mathcal{l}}^{2}+6\mathcal{l}+1-6{m}^{2}=0$, then the equation of circle for which $\mathcal{l}$x + my + 1 = 0 is a tangent will be ?

Hi,
Let centre of circle : (a , b)
a/c to question,
lx + my + 1 is tangent of unknown circle .
so,
radius = distance between centre and touching point .
r = |la + mb + 1|/√(a² + b²)
take square both sides,
r²(a² + b²) = (la + mb + 1)²
r²a² + r²b² = l²a² + m²b² + 1² + 2lmab+2mb + 2la
=> l²(a² - r²) + m²( b² - r²) + 2lmab + 2mb + 2la + 1
now,
given equation ,
3l² + 6l + 1 - 6m² = 0 comapre both equation ,
we get ,
3 = a² - r² -------(1)
-6 = b² - r² ---------(2)
2ab = 0
2b = 0 => b = 0
2a = 6 => a = 3

put value of a in (1)
3 = (3)² - r²
r² = 6
r = √6
hence , centre of circle ( 3, 0) and radius √6

now, equation of circle :
(x -3)² + (y+0)² = √6²
x² - 6x + 9 + y² = 6
x² -6x + y² + 3 = 0

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