The tangent at to the curves touches the circle at
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: D
Solution
The curve is x² = y - 6, so differentiating: 2x = y', giving slope at (1,7) as 2. The tangent line: y - 7 = 2(x - 1) → 2x - y + 5 = 0.
The circle x² + y² + 16x + 12y + c = 0 has center (-8, -6). For the tangent 2x - y + 5 = 0 to touch this circle, the foot of perpendicular from (-8, -6) to the line is the point of tangency.
Using the foot formula: the foot from point (x₀,y₀) to line ax+by+c=0 is:
x = x₀ - a(ax₀+by₀+c)/(a²+b²), y = y₀ - b(ax₀+by₀+c)/(a²+b²)
Here a=2, b=-1, c=5, (x₀,y₀)=(-8,-6):
ax₀+by₀+c = 2(-8)+(-1)(-6)+5 = -16+6+5 = -5
a²+b² = 5
x = -8 - 2(-5)/5 = -8+2 = -6
y = -6 - (-1)(-5)/5 = -6-1 = -7
The point of tangency is (-6, -7).
The circle x² + y² + 16x + 12y + c = 0 has center (-8, -6). For the tangent 2x - y + 5 = 0 to touch this circle, the foot of perpendicular from (-8, -6) to the line is the point of tangency.
Using the foot formula: the foot from point (x₀,y₀) to line ax+by+c=0 is:
x = x₀ - a(ax₀+by₀+c)/(a²+b²), y = y₀ - b(ax₀+by₀+c)/(a²+b²)
Here a=2, b=-1, c=5, (x₀,y₀)=(-8,-6):
ax₀+by₀+c = 2(-8)+(-1)(-6)+5 = -16+6+5 = -5
a²+b² = 5
x = -8 - 2(-5)/5 = -8+2 = -6
y = -6 - (-1)(-5)/5 = -6-1 = -7
The point of tangency is (-6, -7).
[AIEEE 2005]