Multiple choice

The number of common tangents to the circles $x^{2} + y^{2} - 6x - 14y + 48 = 0$ and $x^{2} + y^{2} - 6x = 0$ is

  1. $1$
  2. $2$
  3. $0$
  4. $4$
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D Correct answer
Explanation

Circle 1: x^2 + y^2 - 6x - 14y + 48 = 0. Center (3, 7), r1 = sqrt(9 + 49 - 48) = sqrt(10). Circle 2: x^2 + y^2 - 6x = 0. Center (3, 0), r2 = 3. Distance between centers = sqrt((3-3)^2 + (7-0)^2) = 7. Since 7 > sqrt(10) + 3, the circles are separate, implying 4 common tangents.

AI explanation

Completing the square for the first circle gives (x - 3)^2 + (y - 7)^2 = 10, so its centre is C1(3, 7) and radius is sqrt(10). The second circle is (x - 3)^2 + y^2 = 9, so its centre is C2(3, 0) and radius is 3. The distance between the centres is d = sqrt((3-3)^2 + (7-0)^2) = 7, and the sum of the radii is r1 + r2 = sqrt(10) + 3, which is approximately 6.16. Because the distance between centres (7) is strictly greater than the sum of the radii, the two circles lie completely outside each other, meaning there are 4 common tangents.

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