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If the Equation of a Circle is λX2 + (2λ − 3) Y2 − 4x + 6y − 1 = 0, Then the Coordinates of Centre Are - Mathematics

MCQ

If the equation of a circle is λx2 + (2λ − 3) y2 − 4x + 6y − 1 = 0, then the coordinates of centre are

Options

  • (4/3, −1)

  • (2/3, −1)

  • (−2/3, 1)

  • (2/3, 1)

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Solution

(2/3, −1)

To find the centre:
Coefficient of x2 = Coefficient of y2

\[\therefore \lambda = 2\lambda - 3 \Rightarrow \lambda = 3\]

Therefore, the given equation can be rewritten as

\[3 x^2 + 3 y^2 - 4x + 6y - 1 = 0\].
\[\therefore x^2 + y^2 - \frac{4}{3}x + 2y - \frac{1}{3} = 0\]
Thus, the coordinates of the centre is \[\left( \frac{2}{3}, - 1 \right)\] .
 
Concept: Circle - Standard Equation of a Circle
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 24 The circle
Q 1 | Page 39
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