A Circle with s=0 Touching Another Circle
This problem involves finding a circle with an equation of the form x² + y² + 2gx + 2fy + c = 0, where s = 0 and it touches another circle given by the equation x² + y² - 4x + 6y - 23 = 0.
Understanding the Concepts
- Circle Equation: The general equation of a circle is x² + y² + 2gx + 2fy + c = 0, where (-g, -f) is the center of the circle and √(g² + f² - c) is the radius.
- Condition for Touching: Two circles touch each other if the distance between their centers is equal to the sum or difference of their radii.
- s = 0: This condition implies that the circle passes through the origin (0, 0).
Solving the Problem
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Find the center and radius of the given circle:
- Completing the square for the equation x² + y² - 4x + 6y - 23 = 0, we get:
- (x - 2)² + (y + 3)² = 36
- Therefore, the center of the given circle is (2, -3) and its radius is √36 = 6.
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Use the condition s = 0:
- Since the circle passes through the origin (0, 0), we substitute x = 0 and y = 0 into the general equation of a circle:
- 0² + 0² + 2g(0) + 2f(0) + c = 0
- This simplifies to c = 0.
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Find the center of the required circle:
- Let the center of the required circle be (-g, -f). Since the circles touch, the distance between their centers is equal to the difference of their radii:
- √[(2 + g)² + (-3 + f)²] = 6 - √(g² + f² - c)
- Substituting c = 0, we get:
- √[(2 + g)² + (-3 + f)²] = 6 - √(g² + f²)
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Solve for g and f:
- Squaring both sides of the equation and simplifying, we get:
- 8g - 6f = 32
- We can use the condition c = 0 and the fact that the circle passes through the origin to get another equation:
- g² + f² = 0
- Solving these two equations simultaneously, we find:
- g = 2 and f = -3
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Write the equation of the circle:
- Substituting the values of g, f, and c into the general equation of a circle, we get:
- x² + y² + 4x - 6y = 0
**Therefore, the equation of the circle that touches the given circle and passes through the origin is x² + y² + 4x - 6y = 0.
