Multiple choice

Equation of the circle which is such that the lengths of the tangents to it from the points $( 1,0 ) , ( 0,2 )$ and $( 3,2 )$ are $1 , \sqrt { 7 }$ and $\sqrt { 2 }$ respectively is

  1. $6 \left( x ^ { 2 } + y ^ { 2 } \right) - 28 x - 5 y + 28 = 0$
  2. $9 \left( x ^ { 2 } + y ^ { 2 } \right) - 28 x - 5 y + 28 = 0$
  3. $3 \left( x ^ { 2 } + y ^ { 2 } \right) - 28 x - 5 y + 28 = 0$
  4. $x ^ { 2 } + y ^ { 2 } - 28 x - 5 y + 28 = 0$
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A Correct answer
Explanation

Length of tangent from (x1, y1) to circle x^2+y^2+2gx+2fy+c=0 is sqrt(x1^2+y1^2+2gx1+2fy1+c). Square the lengths: 1 = 1+2g+c, 7 = 4+4f+c, 2 = 9+4+6g+4f+c. Solve the system: 2g+c=0, 4f+c=3, 6g+4f+c=-11. From 1st, c=-2g. From 2nd, 4f-2g=3. From 3rd, 6g+4f-2g=-11 => 4g+4f=-11. Subtract: 6g=-14 => g=-7/3. Then c=14/3. 4f=3+2g=3-14/3=-5/3 => f=-5/12. Equation: x^2+y^2-14/3x-5/6y+14/3=0. Multiply by 6: 6(x^2+y^2)-28x-5y+28=0.

AI explanation

Using the standard formula for the length of a tangent from an external point (x1, y1) to a circle (S = x^2 + y^2 + 2gx + 2fy + c = 0), we know the length squared equals S1. Substituting the given points and squared tangent lengths gives three equations: for (1, 0) we get 1 + 2g + c = 1; for (0, 2) we get 4 + 4f + c = 7; for (3, 2) we get 13 + 6g + 4f + c = 2. Solving this system of equations yields g = -7/3, f = -5/12, and c = 14/3. Substituting these values back into the general circle equation and multiplying the entire equation by 6 to clear the fractions results in 6(x^2 + y^2) - 28x - 5y + 28 = 0.

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