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Difficulty: MediumSystems of Linear and Non-Linear Equations

A circle in the standard (x,y)(x, y) coordinate plane is described by the equation x2+y2=25x^2 + y^2 = 25. A line is described by the equation 3x−4y=c3x - 4y = c, where cc is a positive constant. If the system of these two equations has exactly one real solution for (x,y)(x, y), what is the value of cc?

Answer: 25

Answer

The value of cc is 2525.
The correct answer is 2525. The equation x2+y2=25x^2 + y^2 = 25 represents a circle centered at (0,0)(0, 0) with a radius of 55. For the system of equations to have exactly one real solution, the line 3x−4y=c3x - 4y = c must be tangent to the circle. The perpendicular distance from the center (0,0)(0,0) to the line 3x−4y−c=03x - 4y - c = 0 is given by ∣3(0)−4(0)−c∣32+(−4)2=∣c∣5\frac{|3(0) - 4(0) - c|}{\sqrt{3^2 + (-4)^2}} = \frac{|c|}{5}. Setting this distance equal to the radius of the circle yields ∣c∣5=5\frac{|c|}{5} = 5, which gives ∣c∣=25|c| = 25. Since cc is specified as a positive constant, cc must be 2525.

Step-by-Step Solution

1
Find the center and radius of the circle.
Center is (0,0)(0, 0) and radius is r=5r = 5.
The circle equation x2+y2=25x^2 + y^2 = 25 is in the standard form x2+y2=r2x^2 + y^2 = r^2 centered at the origin with radius r=25=5r = \sqrt{25} = 5.
2
Set up the condition for tangency (exactly one real solution).
The perpendicular distance from the center (0,0)(0,0) to the line 3x−4y−c=03x - 4y - c = 0 must equal the radius 55.
A line intersects a circle at exactly one point if and only if the line is tangent to the circle.
3
Apply the point-to-line distance formula.
Distance d=∣3(0)−4(0)−c∣32+(−4)2=∣c∣5d = \frac{|3(0) - 4(0) - c|}{\sqrt{3^2 + (-4)^2}} = \frac{|c|}{5}.
The formula for the distance from (x0,y0)(x_0, y_0) to the line Ax+By+C=0Ax + By + C = 0 is d=∣Ax0+By0+C∣A2+B2d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}.
4
Solve for the positive constant cc.
c=25c = 25
Setting the distance ∣c∣5\frac{|c|}{5} equal to the radius 55 gives ∣c∣=25|c| = 25. Since cc is a positive constant, c=25c = 25.

Key Concept

Determining conditions for tangency in a system of linear and circular equations.

Alternative Method

Instead of using the geometric distance formula, the system can be solved algebraically by substitution. Express yy in terms of xx from the linear equation: y=3x−c4y = \frac{3x - c}{4}. Substitute this expression into the circle's equation: x2+(3x−c4)2=25x^2 + \left(\frac{3x - c}{4}\right)^2 = 25. Expand the terms and multiply by 1616 to clear the denominator: 16x2+9x2−6cx+c2=40016x^2 + 9x^2 - 6cx + c^2 = 400, which simplifies to the quadratic equation 25x2−6cx+(c2−400)=025x^2 - 6cx + (c^2 - 400) = 0. For the system to have exactly one solution, this quadratic equation must have a discriminant equal to zero. Calculate the discriminant: D=(−6c)2−4(25)(c2−400)=36c2−100c2+40000=−64c2+40000=0D = (-6c)^2 - 4(25)(c^2 - 400) = 36c^2 - 100c^2 + 40000 = -64c^2 + 40000 = 0. Solving for cc yields 64c2=40000  ⟹  c2=62564c^2 = 40000 \implies c^2 = 625. Since cc is a positive constant, c=25c = 25.
Estimated Time:1m 30s
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Topics

Systems of Linear and Non-Linear Equations

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