# 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 ?

Let centre of circle : (a , b)

radius of circle : r

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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