Let C be the circle x^2+y^2+4x−4y−8=0, and let L be the locus of the intersection point of a pair of tangents to C when the angle between the tangents is 60°. At which point does L touch the line x=6?
Answer and solution
Correct answer: A
Solution
Answer: A — (6, 2)
1. Complete squares: (x+2)²+(y−2)²=16. Centre O=(−2,2), radius=4.
2. From an external point P, equal tangents and equal radii give two congruent right triangles. OP bisects the 60° angle.
3. In either triangle, sin 30°=4/OP. Hence OP=8.
4. The locus is (x+2)²+(y−2)²=64. Substitute x=6: 64+(y−2)²=64, so y=2.
5. The single intersection is (6,2), option A.
Why this method works:
Let T be a point of tangency. The radius OT is perpendicular to PT, making OPT right-angled. The segment from the external point to the centre bisects the angle between equal tangents, so ∠OPT=30°.
Independent check and common trap:
In that right triangle sin 30°=OT/OP=4/OP, hence OP=8. The locus is centred at(−2,2) and its rightmost x-coordinate is−2+8=6. The vertical tangent touches at the same y-coordinate as the centre, giving(6,2).
Apply: Complete the square, then OP=r/sin(θ/2)=4/sin 30°=8. The rightmost point of the locus is (−2+8,2)=(6,2).
Check: In that right triangle sin 30°=OT/OP=4/OP, hence OP=8. The locus is centred at(−2,2) and its rightmost x-coordinate is−2+8=6. The vertical tangent touches at the same y-coordinate as the centre, giving(6,2).
1. Complete squares: (x+2)²+(y−2)²=16. Centre O=(−2,2), radius=4.
2. From an external point P, equal tangents and equal radii give two congruent right triangles. OP bisects the 60° angle.
3. In either triangle, sin 30°=4/OP. Hence OP=8.
4. The locus is (x+2)²+(y−2)²=64. Substitute x=6: 64+(y−2)²=64, so y=2.
5. The single intersection is (6,2), option A.
Why this method works:
Let T be a point of tangency. The radius OT is perpendicular to PT, making OPT right-angled. The segment from the external point to the centre bisects the angle between equal tangents, so ∠OPT=30°.
Independent check and common trap:
In that right triangle sin 30°=OT/OP=4/OP, hence OP=8. The locus is centred at(−2,2) and its rightmost x-coordinate is−2+8=6. The vertical tangent touches at the same y-coordinate as the centre, giving(6,2).
G Strategy shortcut
Spot: Coordinate Geometry and Circles.Apply: Complete the square, then OP=r/sin(θ/2)=4/sin 30°=8. The rightmost point of the locus is (−2+8,2)=(6,2).
Check: In that right triangle sin 30°=OT/OP=4/OP, hence OP=8. The locus is centred at(−2,2) and its rightmost x-coordinate is−2+8=6. The vertical tangent touches at the same y-coordinate as the centre, giving(6,2).
G Strategy: Spot: Coordinate Geometry and Circles. Apply: Complete the square, then OP=r/sin(θ/2)=4/sin 30°=8. The rightmost point of the locus is (−2+8,2)=(6,2). Check: In that right triangle sin 30°=OT/OP=4/OP, hence OP=8. The locus is centred at(−2,2) and its rightmost x-coordinate is−2+8=6. The vertical tangent touches at the same y-coordinate as the centre, giving(6,2).