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JEE Main 2023
Circles
Circle
Easy

Question

Choose the correct statement about two circles whose equations are given below : x 2 + y 2 - 10x - 10y + 41 = 0 x 2 + y 2 - 22x - 10y + 137 = 0

Options

Solution

Key Concepts and Formulas

  • Standard Equation of a Circle: The standard form of a circle's equation is (xh)2+(yk)2=r2(x-h)^2 + (y-k)^2 = r^2, where (h,k)(h, k) is the center and rr is the radius.
  • Distance Formula: The distance dd between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by d=(x2x1)2+(y2y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.
  • Conditions for Intersection of Two Circles: If dd is the distance between the centers of two circles with radii r1r_1 and r2r_2, and d=r1+r2d = r_1 + r_2, then the circles touch externally at one point.

Step-by-Step Solution

1. Analyze the First Circle (S1S_1)

The given equation for the first circle is: S1:x2+y210x10y+41=0S_1: x^2 + y^2 - 10x - 10y + 41 = 0

  • Step 1: Convert to Standard Form.

    • Why? Converting to standard form allows us to easily identify the center and radius of the circle. We complete the square for the xx and yy terms.
    • Working: Group the xx and yy terms: (x210x)+(y210y)+41=0(x^2 - 10x) + (y^2 - 10y) + 41 = 0 Complete the square for the xx terms by adding and subtracting (102)2=25(\frac{-10}{2})^2 = 25: (x210x+25)25+(y210y)+41=0(x^2 - 10x + 25) - 25 + (y^2 - 10y) + 41 = 0 Complete the square for the yy terms by adding and subtracting (102)2=25(\frac{-10}{2})^2 = 25: (x210x+25)25+(y210y+25)25+41=0(x^2 - 10x + 25) - 25 + (y^2 - 10y + 25) - 25 + 41 = 0 Rewrite as squared terms: (x5)2+(y5)22525+41=0(x - 5)^2 + (y - 5)^2 - 25 - 25 + 41 = 0 Simplify: (x5)2+(y5)29=0(x - 5)^2 + (y - 5)^2 - 9 = 0 (x5)2+(y5)2=9(x - 5)^2 + (y - 5)^2 = 9
  • Step 2: Identify Center and Radius.

    • Why? By comparing with the standard form, we can read off the center and radius.
    • Working: From (x5)2+(y5)2=9(x - 5)^2 + (y - 5)^2 = 9, we have: Center C1=(5,5)C_1 = (5, 5) Radius r1=9=3r_1 = \sqrt{9} = 3

2. Analyze the Second Circle (S2S_2)

The given equation for the second circle is: S2:x2+y222x10y+137=0S_2: x^2 + y^2 - 22x - 10y + 137 = 0

  • Step 1: Convert to Standard Form.

    • Why? Converting to standard form allows us to easily identify the center and radius of the circle. We complete the square for the xx and yy terms.
    • Working: Group the xx and yy terms: (x222x)+(y210y)+137=0(x^2 - 22x) + (y^2 - 10y) + 137 = 0 Complete the square for the xx terms by adding and subtracting (222)2=121(\frac{-22}{2})^2 = 121: (x222x+121)121+(y210y)+137=0(x^2 - 22x + 121) - 121 + (y^2 - 10y) + 137 = 0 Complete the square for the yy terms by adding and subtracting (102)2=25(\frac{-10}{2})^2 = 25: (x222x+121)121+(y210y+25)25+137=0(x^2 - 22x + 121) - 121 + (y^2 - 10y + 25) - 25 + 137 = 0 Rewrite as squared terms: (x11)2+(y5)212125+137=0(x - 11)^2 + (y - 5)^2 - 121 - 25 + 137 = 0 Simplify: (x11)2+(y5)29=0(x - 11)^2 + (y - 5)^2 - 9 = 0 (x11)2+(y5)2=9(x - 11)^2 + (y - 5)^2 = 9
  • Step 2: Identify Center and Radius.

    • Why? By comparing with the standard form, we can read off the center and radius.
    • Working: From (x11)2+(y5)2=9(x - 11)^2 + (y - 5)^2 = 9, we have: Center C2=(11,5)C_2 = (11, 5) Radius r2=9=3r_2 = \sqrt{9} = 3

3. Compare Centers and Radii, and Calculate Distance Between Centers

  • Step 1: Compare Centers.

    • Why? This directly addresses option (A).
    • Working: C1=(5,5)C_1 = (5, 5) C2=(11,5)C_2 = (11, 5) Since C1C2C_1 \neq C_2, the circles do not have the same center. So, option (A) is incorrect.
  • Step 2: Calculate the Distance Between Centers (dd).

    • Why? The distance between the centers is crucial for determining how the circles interact.
    • Working: Using the distance formula for C1(5,5)C_1(5, 5) and C2(11,5)C_2(11, 5): d=(115)2+(55)2d = \sqrt{(11 - 5)^2 + (5 - 5)^2} d=(6)2+(0)2d = \sqrt{(6)^2 + (0)^2} d=36d = \sqrt{36} d=6d = 6
  • Step 3: Calculate the Sum of Radii.

    • Why? This value is used in the conditions for circle intersection.
    • Working: r1=3r_1 = 3 and r2=3r_2 = 3 Sum of radii: r1+r2=3+3=6r_1 + r_2 = 3 + 3 = 6

4. Determine the Nature of Intersection

  • Why? By comparing the distance between centers (dd) with the sum of radii (r1+r2r_1 + r_2), we can determine the number of meeting points.

  • Working: We found: d=6d = 6 r1+r2=6r_1 + r_2 = 6

    Since d=r1+r2d = r_1 + r_2 (i.e., 6=66 = 6), the circles touch externally. Therefore, they have only one meeting point.

5. Evaluate the Options and Final Answer

Based on our analysis:

  • (A) circles have same centre: Incorrect.
  • (B) circles have no meeting point: Incorrect.
  • (C) circles have only one meeting point: Correct.
  • (D) circles have two meeting points: Incorrect.

Therefore, the correct statement is that the circles have only one meeting point.

Common Mistakes & Tips

  • Completing the Square: Ensure you add and subtract the values correctly when completing the square to maintain the equation's balance.
  • Distance Formula: Double-check the coordinates and signs when using the distance formula.
  • Intersection Conditions: Remember the conditions for circle intersection to determine the number of meeting points accurately.

Summary

To determine the number of intersection points of two circles, convert their equations to standard form to find their centers and radii. Calculate the distance between their centers and compare it with the sum of their radii. In this problem, the distance between the centers equals the sum of the radii, indicating that the circles touch externally and have only one meeting point.

The final answer is \boxed{circles have only one meeting point}, which corresponds to option (C).

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