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What is the center of the circle passing through (2,1)(-2, 1) and tangent to the line 3x2y6=03x - 2y - 6 = 0 at the point (4,3)(4, 3)? \newline(a) (3,2)(3, -2) \newline(b) (417,27)(\frac{41}{7},-\frac{2}{7}) \newline(c)(27,417)(-\frac{2}{7},\frac{41}{7}) \newline(d) (17,397)(\frac{1}{7},\frac{39}{7})

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Q. What is the center of the circle passing through (2,1)(-2, 1) and tangent to the line 3x2y6=03x - 2y - 6 = 0 at the point (4,3)(4, 3)? \newline(a) (3,2)(3, -2) \newline(b) (417,27)(\frac{41}{7},-\frac{2}{7}) \newline(c)(27,417)(-\frac{2}{7},\frac{41}{7}) \newline(d) (17,397)(\frac{1}{7},\frac{39}{7})
  1. Identify Radius Calculation: Identify the radius of the circle using the tangent point and the line equation. The radius is perpendicular to the tangent line at the point of tangency. Calculate the slope of the line 3x2y6=03x - 2y - 6 = 0.
  2. Equation of Radius Line: Use the point-slope form to find the equation of the line that is the radius of the circle. This line passes through (4,3)(4, 3) and has a slope of 23-\frac{2}{3}.
  3. Simplify Radius Line Equation: Simplify the equation of the radius line.
  4. Midpoint Calculation: Find the midpoint of the line segment joining the center of the circle and the point (4,3)(4, 3) since the radius passes through this midpoint. The other endpoint is (2,1)(-2, 1).
  5. Circle Equation Setup: Set up the equation of the circle with the center at the midpoint and passing through (4,3)(4, 3).

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