Math Problem Statement

A diameter of a circle C1 has endpoints at (-3, -5) and (7, 3). Another circle C2 has center (10, 5) and radius same as C1. The two circles intersect at points R and S. Find an equation of the circle C1. Show that the equation of line RS is y = -2x + 13 and hence that the x-coordinates of R and S satisfy the equation 5x^2 - 60x + 159 = 0. Also, given points A (-7, 3) and B (5, 11), find the equation of the perpendicular bisector of AB, and find the equation of a circle passing through A and B, whose center lies on 12x - 5y = 70.

Solution

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6. A diameter of a circle:

  • (a) Find an equation of the circle C1C_1, which has endpoints at (3,5)(-3, -5) and (7,3)(7, 3).
  • (b) Another circle C2C_2 has center (10,5)(10, 5) and the same radius as C1C_1. Find the equation of the circle C2C_2.
  • (c) The two circles intersect at points RR and SS. Show that the equation of the line RSRS is y=2x+13y = -2x + 13.
  • (d) Show that the xx-coordinates of RR and SS satisfy the equation 5x260x+159=05x^2 - 60x + 159 = 0.

7. The coordinates of two points AA and BB are (7,3)(-7, 3) and (5,11)(5, 11) respectively:

  • (a) Show that the equation of the perpendicular bisector of ABAB is 3x+2y=113x + 2y = 11.
  • (b) A circle passes through AA and BB, and its center lies on the line 12x5y=7012x - 5y = 70. Find the equation of the circle.

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Math Problem Analysis

Mathematical Concepts

Circle Geometry
Line Equations
Simultaneous Equations

Formulas

Equation of a circle: (x - h)^2 + (y - k)^2 = r^2
Distance formula: sqrt((x2 - x1)^2 + (y2 - y1)^2)
Midpoint formula: ((x1 + x2)/2, (y1 + y2)/2)
Slope formula: (y2 - y1) / (x2 - x1)

Theorems

Perpendicular Bisector Theorem
Quadratic Equation Theorem

Suitable Grade Level

Grades 10-12