Question:
The radius of the circle 3x(x − 2) + 3y(y + 1) = 4 is
वृत्त 3x(x − 2) + 3y(y + 1) = 4 की त्रिज्या क्या है?
Options:
- 2
- \(\sqrt{\dfrac{31}{12}}\)
- 3
- \(\sqrt{\dfrac{15}{4}}\)
- 2
- \(\sqrt{\dfrac{31}{12}}\)
- 3
- \(\sqrt{\dfrac{15}{4}}\)
Correct Answer: b)
Explanation:
Given equation: \(3x^2 – 6x + 3y^2 + 3y = 4 \implies x^2 – 2x + y^2 + y = \dfrac{4}{3}\).
Completing squares: \((x-1)^2 + \left(y+\dfrac{1}{2}\right)^2 = \dfrac{4}{3} + 1 + \dfrac{1}{4} = \dfrac{16+12+3}{12} = \dfrac{31}{12}\).
Radius \(r = \sqrt{\dfrac{31}{12}}\).
समीकरण को मानक रूप में लिखने पर: \((x-1)^2 + \left(y+\dfrac{1}{2}\right)^2 = \dfrac{31}{12}\)।
अतः त्रिज्या \(r = \sqrt{\dfrac{31}{12}}\) है।