Question

The circles $x^{2} + y^{2} - 2x - 4y = 0$ and $x^{2} + y^{2} - 8y - 4 = 0$ have

A. 2 common tangents.

B. 3 common tangents.

C. 4 common tangents.

D. 1 common tangent.

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Answer

To address the given question, first we need to analyze the circles described by the equations:

$$ x^{2} + y^{2} - 2x - 4y = 0 \quad \text{and} \quad x^{2} + y^{2} - 8y - 4 = 0 $$

Step 1: Determine the center and radius of each circle.

For the first circle, converting to standard form: $$ (x-1)^2 + (y-2)^2 = 5 $$ Thus, the center is $ C_1 = (1, 2) $ and the radius is $ r_1 = \sqrt{5} $.

For the second circle, converting to standard form: $$ x^2 + (y-4)^2 = 20 $$ Thus, the center is $ C_2 = (0, 4) $ and the radius is $ r_2 = 2\sqrt{5} $.

Step 2: Determine the distance between the centers $ C_1 $ and $ C_2 $. $$ C_1C_2 = \sqrt{(1-0)^2 + (2-4)^2} = \sqrt{1 + 4} = \sqrt{5} $$

Step 3: Compare the distance between the centers with the difference in radii. $$ r_2 - r_1 = 2\sqrt{5} - \sqrt{5} = \sqrt{5} $$ Since $ C_1C_2 = r_2 - r_1 = \sqrt{5} $, the two circles touch internally.

Conclusion: As the circles only touch internally at exactly one point, they will have exactly one common tangent.

Therefore, the correct option is (D) 1 common tangent.


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