Question

If 4a^2-5b^2+6a+1=0 and the line ax+by+1=0 touches a fixed circle, when radius is

Answer 1

Let a circle of radius r and centre: ( α , β) touches the line ax + by + 1 = 0
means distance between centre to circumference of circle = radius of circle
∴ distance between (α, β) and (ax + by + 1 = 0) = r
⇒ |aα + bβ + 1 |/√(a² + b²) = r
Squaring both sides,
⇒ (aα + bβ + 1)² = r² (a² + b²)
⇒ a²α² + b²β² + 1 + 2abαβ + 2bβ + 2aα = r²a² + r²b²
⇒ (α² - r²)a² + (β² - r²)b² + 2abαβ + 2aα + 2bβ + 1 = 0 , compare this equation with given given 4a² - 5b² + 6a + 1 = 0
4/(α² - r²) = -5/(β² - r²) = 0/2ab = 6/2α = 0/2β = 1/1
β = 0 , α = 3
so, 4 = α² - r² ⇒ 4 = 9 - r²
r = √5 unit

Hence, answer is √5 unit