Question

The circles

$$ \begin{array}{l} x^{2}+y^{2}-8 x+6 y+21=0, \ x^{2}+y^{2}+4 x-10 y-115=0 \end{array} $$

are: a) intersecting b) touching externally c) touching internally d) one is lying inside the other

A intersecting B touching externally C touching internally D one is lying inside the other

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Answer

To determine the relationship between the given circles represented by the equations

$$ \begin{align*} x^2 + y^2 - 8x + 6y + 21 &= 0, \ x^2 + y^2 + 4x - 10y - 115 &= 0, \end{align*} $$

we follow these steps:

Step 1: Identify the Centers and Radii

First, we rewrite each equation in standard form for a circle ((x - h)^2 + (y - k)^2 = r^2).

For the first circle:

  • Equation: $x^2 + y^2 - 8x + 6y + 21 = 0$
  • Completing the square for (x) and (y): $$ x^2 - 8x + y^2 + 6y = -21 \ (x - 4)^2 - 16 + (y + 3)^2 - 9 = -21 \ (x - 4)^2 + (y + 3)^2 = 4 $$
  • Center, (\mathbf{C_1}): ((4, -3))
  • Radius, (R_1): (2) (since (4 = 2^2))

For the second circle:

  • Equation: $x^2 + y^2 + 4x - 10y - 115 = 0$
  • Completing the square for (x) and (y): $$ x^2 + 4x + y^2 - 10y = 115 \ (x + 2)^2 - 4 + (y - 5)^2 - 25 = 115 \ (x + 2)^2 + (y - 5)^2 = 144 $$
  • Center, (\mathbf{C_2}): ((-2, 5))
  • Radius, (R_2): (12) (since (144 = 12^2))

Step 2: Calculate the Distance Between Centers

Utilizing the distance formula, the distance between centers (\mathbf{C_1}) and (\mathbf{C_2}) is: $$ \text{Distance } (\mathbf{C_1C_2}) = \sqrt{(4 - (-2))^2 + (-3 - 5)^2} \ = \sqrt{(6)^2 + (-8)^2} \ = \sqrt{36 + 64} \ = \sqrt{100} \ = 10 $$

Step 3: Compare the Distance with Radii Sum and Difference

For the circles to be touching internally: $$\mathbf{|R_2 - R_1| = \mathbf{C_1C_2}}$$ By substitution: $$|12 - 2| = 10$$ This implies the given circles have an internal tangency relationship as (|10| = 10).

Conclusion

Since the distance between the centers of the circles is equal to the difference in their radii, the circles are touching internally.

Correct Answer: Option C


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