Math Problem Statement

Find the equation of the circle which has its center on the line x + y = 0 and passes through the points of intersection of the two circles (x - 1)^2 + (y + 5)^2 = 50 and (x + 1)^2 + (y + 1)^2 = 10.

Solution

We need to solve the problem step by step. Here's the setup:

We are tasked to:

  1. Find the equation of a circle that:
    • Has its center on the line x+y=0x + y = 0,
    • Passes through the points of intersection of the given circles: (x1)2+(y+5)2=50and(x+1)2+(y+1)2=10.(x - 1)^2 + (y + 5)^2 = 50 \quad \text{and} \quad (x + 1)^2 + (y + 1)^2 = 10.

Step 1: Find points of intersection of the two circles

The two circles are: (x1)2+(y+5)2=50(Circle 1),(x - 1)^2 + (y + 5)^2 = 50 \quad \text{(Circle 1)}, (x+1)2+(y+1)2=10(Circle 2).(x + 1)^2 + (y + 1)^2 = 10 \quad \text{(Circle 2)}.

Expanding both equations: For Circle 1: (x^2 - 2x + 1) + (y^2 + 10y + 25) = 50 \implies x^2 + y^2 - 2x + 10y + 26 = 50 \implies x^2 + y^2 - 2x + 10y - 24 = 0. \tag{1}

For Circle 2: (x^2 + 2x + 1) + (y^2 + 2y + 1) = 10 \implies x^2 + y^2 + 2x + 2y + 2 = 10 \implies x^2 + y^2 + 2x + 2y - 8 = 0. \tag{2}

Now, subtract (2)(2) from (1)(1): (x2+y22x+10y24)(x2+y2+2x+2y8)=0.(x^2 + y^2 - 2x + 10y - 24) - (x^2 + y^2 + 2x + 2y - 8) = 0. -4x + 8y - 16 = 0 \implies -x + 2y - 4 = 0 \implies x = 2y - 4. \tag{3}


Step 2: Substitute x=2y4x = 2y - 4 into one circle equation

Using (3)(3) in (1)(1): (2y4)2+y22(2y4)+10y24=0.(2y - 4)^2 + y^2 - 2(2y - 4) + 10y - 24 = 0. Expand and simplify: (4y216y+16)+y24y+8+10y24=0.(4y^2 - 16y + 16) + y^2 - 4y + 8 + 10y - 24 = 0. 5y210y0=0    y(y2)=0.5y^2 - 10y - 0 = 0 \implies y(y - 2) = 0. So, y=0y = 0 or y=2y = 2.

If y=0y = 0, then x=2(0)4=4x = 2(0) - 4 = -4. If y=2y = 2, then x=2(2)4=0x = 2(2) - 4 = 0.

Thus, the points of intersection are: (4,0)and(0,2).(-4, 0) \quad \text{and} \quad (0, 2).


Step 3: Equation of the required circle

The circle passes through (4,0)(-4, 0) and (0,2)(0, 2) and has its center on the line x+y=0x + y = 0. Let the center of the circle be (h,k)(h, k), where h+k=0h + k = 0, so k=hk = -h.

The equation of the circle is: (xh)2+(yk)2=r2.(x - h)^2 + (y - k)^2 = r^2. Substitute k=hk = -h: (x - h)^2 + (y + h)^2 = r^2. \tag{4}


Step 4: Use the given points to find hh and r2r^2

Substitute (4,0)(-4, 0) into (4)(4): (4h)2+(0+h)2=r2.(-4 - h)^2 + (0 + h)^2 = r^2. (16 + 8h + h^2) + h^2 = r^2 \implies 2h^2 + 8h + 16 = r^2. \tag{5}

Substitute (0,2)(0, 2) into (4)(4): (0h)2+(2+h)2=r2.(0 - h)^2 + (2 + h)^2 = r^2. h^2 + (4 + 4h + h^2) = r^2 \implies 2h^2 + 4h + 4 = r^2. \tag{6}

From (5)(5) and (6)(6): 2h2+8h+16=2h2+4h+4.2h^2 + 8h + 16 = 2h^2 + 4h + 4. 8h+16=4h+4    4h=12    h=3.8h + 16 = 4h + 4 \implies 4h = -12 \implies h = -3.

Thus, k=h=3k = -h = 3, and the center is (3,3)(-3, 3).


Step 5: Find r2r^2

Substitute h=3h = -3 into (6)(6): 2(3)2+4(3)+4=r2.2(-3)^2 + 4(-3) + 4 = r^2. 2(9)12+4=r2    1812+4=r2    r2=10.2(9) - 12 + 4 = r^2 \implies 18 - 12 + 4 = r^2 \implies r^2 = 10.


Final Answer:

The equation of the required circle is: (x+3)2+(y3)2=10.(x + 3)^2 + (y - 3)^2 = 10.


Would you like a detailed explanation for any step?

Related Questions:

  1. How do you find the equation of a circle passing through three points?
  2. How are the points of intersection of two circles determined geometrically?
  3. What is the significance of a circle’s center lying on a line?
  4. How can substitution help in solving simultaneous equations of conic sections?
  5. What other properties of circles can be derived from their equations?

Tip:

Always simplify the circle equations before solving intersection problems to reduce computational complexity.

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Math Problem Analysis

Mathematical Concepts

Coordinate Geometry
Circles
Simultaneous Equations
Algebraic Substitution

Formulas

Equation of a circle: (x - h)^2 + (y - k)^2 = r^2
Expanding squared terms: (x ± a)^2 = x^2 ± 2ax + a^2
Solving simultaneous equations: Substitution and elimination

Theorems

Properties of Circles
Intersection of Two Circles

Suitable Grade Level

Grade 11-12