The diameter of the circle 9 x 2- y 2-4 X 2- Y 2-8 X isA- 4 - 5B- 8 - 5C- 8D- Does not exit

The correct option is B 8/5 9x^2+y^2 = 4(x^2 y^2) 8x. 5x^2+5y^2+8x=0 x^2+y^2+8/5x=0 g= 4/5 Radius =√(g^2+f^2 c) =4/5 Diameter =2 ×4/5 =8/5 ...

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Standard XII

Mathematics

Definition of Circle

The diameter ...

Question

The diameter of the circle $9 x^{2} + y^{2}$ = $4 (X^{2} - Y^{2}) - 8 X$ is

4/5

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8/5

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Solution

The correct option is B
8/5

$9 x^{2} + y^{2}$ = $4 (x^{2} - y^{2}) - 8 x .$
$5 x^{2} + 5 y^{2} + 8 x = 0$
$x^{2} + y^{2} + \frac{8}{5} x = 0$
$g = \frac{- 4}{5}$
Radius $= \sqrt{g^{2} + f^{2} - c}$
$= \frac{4}{5}$
Diameter $= 2 \times \frac{4}{5}$
$= \frac{8}{5}$

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The diameter of the circle 9x2+y2 = 4(X2−Y2)−8X is

The correct option is B 8/5 9x2+y2 = 4(x2−y2)−8x. 5x2+5y2+8x=0 x2+y2+85x=0 g=−45 Radius =√g2+f2−c =45 Diameter =2×45 =85

The diameter of the circle $9 x^{2} + y^{2}$ = $4 (X^{2} - Y^{2}) - 8 X$ is

The correct option is B
8/5

$9 x^{2} + y^{2}$ = $4 (x^{2} - y^{2}) - 8 x .$
$5 x^{2} + 5 y^{2} + 8 x = 0$
$x^{2} + y^{2} + \frac{8}{5} x = 0$
$g = \frac{- 4}{5}$
Radius $= \sqrt{g^{2} + f^{2} - c}$
$= \frac{4}{5}$
Diameter $= 2 \times \frac{4}{5}$
$= \frac{8}{5}$