Math Problem Statement

find the equation of the circle of radius √5 tangent to the line 2x+y-3=0 and having its center on the line 2x+3y=4

Solution

To find the equation of the circle with radius 5\sqrt{5}, tangent to the line 2x+y3=02x + y - 3 = 0 and having its center on the line 2x+3y=42x + 3y = 4, we can follow these steps:

  1. Identify the line equations:

    • The first line (the tangent line) is 2x+y3=02x + y - 3 = 0.
    • The second line (on which the center lies) is 2x+3y4=02x + 3y - 4 = 0.
  2. Find the distance from the center of the circle to the tangent line: The distance dd from a point (x0,y0)(x_0, y_0) to the line Ax+By+C=0Ax + By + C = 0 is given by: d=Ax0+By0+CA2+B2d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} For the line 2x+y3=02x + y - 3 = 0, A=2A = 2, B=1B = 1, and C=3C = -3. The radius of the circle is 5\sqrt{5}.

    Therefore, we have: 2x0+y0322+12=5\frac{|2x_0 + y_0 - 3|}{\sqrt{2^2 + 1^2}} = \sqrt{5} Simplifying, we find: 2x0+y03=55=5|2x_0 + y_0 - 3| = \sqrt{5} \cdot \sqrt{5} = 5 Thus: 2x0+y03=5or2x0+y03=52x_0 + y_0 - 3 = 5 \quad \text{or} \quad 2x_0 + y_0 - 3 = -5 This leads to two cases:

    • Case 1: 2x0+y0=82x_0 + y_0 = 8
    • Case 2: 2x0+y0=22x_0 + y_0 = -2
  3. Substitute into the second line equation: The center (x0,y0)(x_0, y_0) must also satisfy 2x0+3y0=42x_0 + 3y_0 = 4.

    For Case 1:

    • From 2x0+y0=82x_0 + y_0 = 8, we can express y0y_0 as y0=82x0y_0 = 8 - 2x_0.
    • Substituting into the second line equation: 2x0+3(82x0)=42x_0 + 3(8 - 2x_0) = 4 2x0+246x0=4    4x0+24=4    4x0=20    x0=52x_0 + 24 - 6x_0 = 4 \implies -4x_0 + 24 = 4 \implies -4x_0 = -20 \implies x_0 = 5
    • Then: y0=82(5)=2y_0 = 8 - 2(5) = -2
    • So, the center is (5,2)(5, -2).

    For Case 2:

    • From 2x0+y0=22x_0 + y_0 = -2, we have y0=22x0y_0 = -2 - 2x_0.
    • Substituting into the second line equation: 2x0+3(22x0)=42x_0 + 3(-2 - 2x_0) = 4 2x066x0=4    4x06=4    4x0=10    x0=522x_0 - 6 - 6x_0 = 4 \implies -4x_0 - 6 = 4 \implies -4x_0 = 10 \implies x_0 = -\frac{5}{2}
    • Then: y0=22(52)=2+5=3y_0 = -2 - 2\left(-\frac{5}{2}\right) = -2 + 5 = 3
    • So, the center is (52,3)\left(-\frac{5}{2}, 3\right).
  4. Find the equations of the circles: The equation of a circle with center (h,k)(h, k) and radius rr is given by: (xh)2+(yk)2=r2(x - h)^2 + (y - k)^2 = r^2

    • For center (5,2)(5, -2): (x5)2+(y+2)2=5(x - 5)^2 + (y + 2)^2 = 5
    • For center (52,3)\left(-\frac{5}{2}, 3\right): (x+52)2+(y3)2=5\left(x + \frac{5}{2}\right)^2 + (y - 3)^2 = 5

Thus, the two possible equations of the circles are:

  1. (x5)2+(y+2)2=5(x - 5)^2 + (y + 2)^2 = 5
  2. (x+52)2+(y3)2=5\left(x + \frac{5}{2}\right)^2 + (y - 3)^2 = 5

Would you like any further details or have questions about this problem? Here are some related questions for further exploration:

  1. How can we find the point of tangency on the line?
  2. What is the significance of the center being on a specific line?
  3. How does the radius affect the position of the circle relative to the tangent line?
  4. Can you derive the slope of the tangent line?
  5. How do you determine if two circles intersect or are tangent?

Tip: Always check the conditions of tangency and distance when dealing with circles and lines!

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Math Problem Analysis

Mathematical Concepts

Geometry
Circles
Linear Equations

Formulas

Distance from point to line: d = |Ax_0 + By_0 + C| / √(A^2 + B^2)
Circle equation: (x - h)^2 + (y - k)^2 = r^2

Theorems

The relationship between tangents and circles
Distance formula

Suitable Grade Level

Grades 10-12