Multiple choice

The tangent to the circle ${x^2} + {y^2} = 5$ at the point $\left( {1, - 2} \right)$ also touches the circle ${x^2} + {y^2} - 8x + 6y + 20 = 0$ at

  1. $\left( { - 2,1} \right)$
  2. $\left( { - 3,0} \right)$
  3. $\left( { - 1, - 1} \right)$
  4. $\left( { 3, - 1} \right)$
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D Correct answer
Explanation

Tangent to x^2 + y^2 = 5 at (1, -2) is x - 2y = 5, or x = 2y + 5. Substitute into the second circle: (2y+5)^2 + y^2 - 8(2y+5) + 6y + 20 = 0. 4y^2 + 20y + 25 + y^2 - 16y - 40 + 6y + 20 = 0. 5y^2 + 10y + 5 = 0. y^2 + 2y + 1 = 0, so (y+1)^2 = 0, y = -1. If y = -1, x = 2(-1) + 5 = 3. Point is (3, -1).

AI explanation

The center of the second circle is (4, -3) and its radius is the square root of (16 plus 9 minus 20), which equals the square root of 5. The tangent to the first circle at (1, -2) has the equation 1x minus 2y equals 5. The distance from the center (4, -3) to this tangent line is the absolute value of (4 plus 6 minus 5) divided by the square root of 5, which equals the square root of 5, confirming the line is tangent to the second circle. Solving the system of equations for the second circle and the tangent line x minus 2y equals 5 yields the point of tangency as (3, -1).

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