Math Problem Statement

Find the equation of the circle which passes through the point (1, 2) and cuts orthogonally each of the circles x^2 + y^2 = 9 and x^2 + y^2 − 2x + 8y − 7 = 0.

Solution

To solve this problem, we need to find the equation of the circle that passes through the point (1,2)(1, 2) and cuts orthogonally through the given circles:

  1. x2+y2=9x^2 + y^2 = 9
  2. x2+y22x+8y7=0x^2 + y^2 - 2x + 8y - 7 = 0

General Strategy:

  1. Equation of the required circle: The general equation of a circle is: x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0 where gg, ff, and cc are constants to be determined.

  2. Condition for orthogonality: Two circles are orthogonal if the product of their gradients at any point of intersection is 1-1. Algebraically, for two circles to cut orthogonally, the following condition must hold: 2g1g2+2f1f2=c1+c22g_1g_2 + 2f_1f_2 = c_1 + c_2 where g1g_1, f1f_1, and c1c_1 are the constants from the equation of the first circle, and g2g_2, f2f_2, and c2c_2 are the constants from the second circle.

  3. First circle: For the circle x2+y2=9x^2 + y^2 = 9, we compare this with the general form x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0, giving:

    • g1=0g_1 = 0, f1=0f_1 = 0, and c1=9c_1 = -9.
  4. Second circle: For the circle x2+y22x+8y7=0x^2 + y^2 - 2x + 8y - 7 = 0, this matches the general form, giving:

    • g2=1g_2 = -1, f2=4f_2 = 4, and c2=7c_2 = -7.

Step-by-Step Solution:

  1. Orthogonality condition with the first circle: Using the condition 2g1g+2f1f=c1+c2g_1g + 2f_1f = c_1 + c, we substitute the values: 2(0)(g)+2(0)(f)=9+c2(0)(g) + 2(0)(f) = -9 + c Simplifying, we get: c=9c = 9

  2. Orthogonality condition with the second circle: Using the condition 2g2g+2f2f=c2+c2g_2g + 2f_2f = c_2 + c, substitute the values: 2(1)(g)+2(4)(f)=7+92(-1)(g) + 2(4)(f) = -7 + 9 Simplifying this: 2g+8f=2-2g + 8f = 2 g+4f=1(Equation 1)-g + 4f = 1 \quad \text{(Equation 1)}

  3. Passing through the point (1,2)(1, 2): The required circle passes through the point (1,2)(1, 2), so substitute x=1x = 1 and y=2y = 2 into the equation x2+y2+2gx+2fy+c=0x^2 + y^2 + 2gx + 2fy + c = 0: 12+22+2g(1)+2f(2)+9=01^2 + 2^2 + 2g(1) + 2f(2) + 9 = 0 Simplifying: 1+4+2g+4f+9=01 + 4 + 2g + 4f + 9 = 0 2g+4f=142g + 4f = -14 g+2f=7(Equation 2)g + 2f = -7 \quad \text{(Equation 2)}

  4. Solving the system of equations: We have the system of equations:

    1. g+4f=1-g + 4f = 1
    2. g+2f=7g + 2f = -7

    Adding both equations: (g+g)+(4f+2f)=17(-g + g) + (4f + 2f) = 1 - 7 6f=66f = -6 f=1f = -1 Substituting f=1f = -1 into Equation 2: g+2(1)=7g + 2(-1) = -7 g2=7g - 2 = -7 g=5g = -5

Final Equation:

Now that we have g=5g = -5, f=1f = -1, and c=9c = 9, the equation of the required circle is: x2+y210x2y+9=0x^2 + y^2 - 10x - 2y + 9 = 0

Would you like more details on any part of the solution? Here are five related questions for further exploration:

  1. What is the geometric interpretation of two circles cutting orthogonally?
  2. How can the condition for orthogonality be generalized for different conic sections?
  3. Why is the constant cc in the circle equation important in determining its size?
  4. How would the equation change if the circle were tangent instead of orthogonal to the others?
  5. Can you derive the equation of a circle given three points on its circumference?

Tip: Always cross-check conditions such as orthogonality with the general properties of geometric figures to ensure accurate results!

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Math Problem Analysis

Mathematical Concepts

Circle Equation
Orthogonal Circles
Algebraic Geometry

Formulas

General equation of a circle: x^2 + y^2 + 2gx + 2fy + c = 0
Condition for orthogonality: 2g1g2 + 2f1f2 = c1 + c2

Theorems

Orthogonality of Circles

Suitable Grade Level

Grades 11-12 / College level