If the equation 4 a-3 x2-a y2-6 x-2 y-2-0 represents a circle- then its centre isa 3-1b 3-1c -3-1d none of these

(c) ( minus;3, 1) If the equation (4a minus; 3) x2 + ay2 + 6x minus; 2y + 2 = 0 represents a circle, then we have: Coefficient of x2 = Coefficient of y ...

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Standard XI

Mathematics

Point Slope Form of a Line

If the equati...

Question

If the equation (4a − 3) x² + ay² + 6x − 2y + 2 = 0 represents a circle, then its centre is
(a) (3, −1)
(b) (3, 1)
(c) (−3, 1)
(d) none of these

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Solution

(c) (−3, 1)

If the equation (4a − 3) x² + ay² + 6x − 2y + 2 = 0 represents a circle, then we have:
Coefficient of x² = Coefficient of y²
⇒ $4 a - 3 = a$
⇒ a = 1

∴ Equation of the circle = $x^{2} + y^{2} + 6 x - 2 y + 2 = 0$

Thus, the coordinates of the centre is $(- 3, 1)$ .

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If the equation (4a − 3) x2 + ay2 + 6x − 2y + 2 = 0 represents a circle, then its centre is (a) (3, −1) (b) (3, 1) (c) (−3, 1) (d) none of these

(c) (−3, 1) If the equation (4a − 3) x2 + ay2 + 6x − 2y + 2 = 0 represents a circle, then we have: Coefficient of x2 = Coefficient of y2 ⇒ 4a-3=a ⇒ a = 1 ∴ Equation of the circle = x2+y2+6x-2y+2=0 Thus, the coordinates of the centre is -3, 1.

If the equation (4a − 3) x² + ay² + 6x − 2y + 2 = 0 represents a circle, then its centre is
(a) (3, −1)
(b) (3, 1)
(c) (−3, 1)
(d) none of these

(c) (−3, 1)

If the equation (4a − 3) x² + ay² + 6x − 2y + 2 = 0 represents a circle, then we have:
Coefficient of x² = Coefficient of y²
⇒ $4 a - 3 = a$
⇒ a = 1

∴ Equation of the circle = $x^{2} + y^{2} + 6 x - 2 y + 2 = 0$

Thus, the coordinates of the centre is $(- 3, 1)$ .