Question

Help! 50 points. Class 11! If 3l^2 + 6l + 1 - 6m^2 = 0, then find the equation of circle for which lx+my+1=0 is a tangent.

Answer 1

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 . ( see the attachment )
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

Answer 2

Answer:

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