Multiple choice

The point from which the lengths of tangents to the three circles $\mathrm{x}^{2}+\mathrm{y}^{2}-4=0,\ \mathrm{x}^{2}+\mathrm{y}^{2}$ -2x $+3\mathrm{y}=0$ and $\mathrm{x}^{2}+\mathrm{y}^{2}+7\mathrm{y}-18=0$ are equal is

  1. $(2, 5)$
  2. $(3, 4)$
  3. $(4, 3)$
  4. $(5, 2)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The point (x, y) from which tangent lengths are equal is the radical center of the three circles. The radical axis of the first two circles is x^2+y^2-4 = x^2+y^2-2x+3y => 2x-3y=4. Testing (5, 2): 2(5)-3(2) = 10-6 = 4. This satisfies the first radical axis. Checking the third circle: x^2+y^2+7y-18 = x^2+y^2-4 => 7y-18 = -4 => 7y=14 => y=2. If y=2, 2x-3(2)=4 => 2x=10 => x=5.

AI explanation

The point from which the tangent lengths to three circles are equal is their radical centre, found by equating the circle equations. Subtracting the second equation from the first gives 2x - 3y + 4 = 0, and subtracting the third from the first gives -7y + 14 = 0, which simplifies to y = 2. Substituting y = 2 into the first equation yields 2x - 6 + 4 = 0, solving to x = 5, so the coordinates are (5, 2).

Keep practicing — related questions