If one of the diameters of the circle x 2 + y 2 − 2 2 x − 6 2 y + 1 4 = 0 is a chord of circle. ( x − 2 2 ) 2 + ( y − 2 2 ) 2 = r 2 , then the value of r2 is equal to…………….

x 2 + y 2 − 2 2 x − 6 2 y + 1 4 = 0 centre ( 2 , 3 2 ) radius ( ( 2 ) 2 + ( 3 2 ) 2 − 1 4 ) 1 / 2 = ( 2 + 1 8 − 1 4 ) 1 / 2 = ( 6 ) ( x − 2 2 ) 2 + ( y − 2 2 ) 2 = r 2 centre ( 2 2 , 2 2 ) OA = ( 2 2 − 2 ) 2 + ( 2 2 − 3 2 ) 2 = 2 + 2 = 2 r2 = ( 6 ) 2 + ( 2 ) 2 = 6 + 4 = 1 0

If one of the diameters of the circle $x^{2} + y^{2} - 2 \sqrt[]{2} x - 6 \sqrt[]{2} y + 14 = 0$ is a chord of circle. ${(x - 2 \sqrt[]{2})}^{2} + {(y - 2 \sqrt[]{2})}^{2} = r^{2},$ then the value of r² is equal to…………….

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Alok Kumar Singh | Contributor-Level 10

6 months ago

$x^{2} + y^{2} - 2 \sqrt[]{2} x - 6 \sqrt[]{2} y + 14 = 0$

centre $(\sqrt[]{2}, 3 \sqrt[]{2})$

radius ${({(\sqrt[]{2})}^{2} + {(3 \sqrt[]{2})}^{2} - 14)}^{1 / 2}$

$= {(2 + 18 - 14)}^{1 / 2} = (\sqrt[]{6})$

${(x - 2 \sqrt[]{2})}^{2} + {(y - 2 \sqrt[]{2})}^{2} = r^{2}$

centre $(2 \sqrt[]{2}, 2 \sqrt[]{2})$

OA = $\sqrt[]{{(2 \sqrt[]{2} - \sqrt[]{2})}^{2} + {(2 \sqrt[]{2} - 3 \sqrt[]{2})}^{2}} = \sqrt[]{2 + 2} = 2$

r² = ${(\sqrt[]{6})}^{2} + {(2)}^{2} = 6 + 4 = 10$

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(i) If $- 4 x \geq 12$ , then $x \dots - 3$ .

(ii) If $\frac{3}{4} - x \leq - 3,$ then $x \dots 4$ .

(iii) If $\frac{2}{x + 2} > 0,$ then $x \dots - 2$ .

(iv) If $x > - 5$ , then $4 x \dots - 20$ .

(v) If $x > y$ and $z < 0$ , then $- x z \dots - y z$ .

(vi) If $p > 0$ and $q < 0$ , then $p - q \dots p$ .

(vii) If $x + 2 > 5$ , then $x \dots - 7$ or $x \dots 3$ .

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(i) If $x < y$ and $b < 0$ , then $\frac{x}{b} > \frac{y}{b} .$

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(xi) Graph of $x \geq 0$ corresponds to Fig 6.13.

(xii) Graph of $y \leq 0$ corresponds to Fig 6.14.

(xiii) Solution set of $x \geq 0$ and $y \leq 0$ corresponds to Fig 6.15.

(xiv) Solution set of $x \geq 0$ and $y \leq 1$ corresponds to Fig 6.16.

(xv) Solution set of $x + y \geq 0$ corresponds to Fig 6.17.

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