QBManager

Conic Section

119646 The number of integral values of $λ$ for which the equation $x^{2} + y^{2} - 2 λ x + 2 λ y + 14 = 0$ represents a circle whose radius cannot exceed 6 is

1 10

2 11

3 12

4 9

Explanation:

B $x^{2} + y^{2} - 2 λ x + 2 λ y + 14 = 0$
radius $= \sqrt{λ^{2} + λ^{2} - 14}$
We have radius $\leq 6$ .
(radius) $^{2} \leq 36$
$2 λ^{2} - 14 \leq 36$
$2 λ^{2} \leq 50$
$λ^{2} \leq 25$
$λ^{2} - 25 \leq 0$
$- 5 \leq λ \leq 5$
So, number of integral values of $λ$ is 11 .
$λ = - 5, - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, 5$

BCECE-2017

Conic Section

119630 Area of the circle in which a chord of length $\sqrt{2}$ makes an angle $π / 2$ at the centre, is

1

π / 2

squnits

2

2 π

sq units

3

π

sq units

4

π / 4

squnits

Explanation:

C :
original image
Given, chord $(A B) = \sqrt{2}$
$θ = \frac{π}{2}$
Radius of circle $(r) = \frac{A B / 2}{\sin \frac{θ}{2}} = \frac{\sqrt{2} / 2}{\sin \frac{π / 2}{2}}$
$r = \frac{1 / \sqrt{2}}{1 / \sqrt{2}} = 1 unit$
Area of the circle $= π r^{2}$
Area of the circle $= π \times 1^{2}$
Area of the circle $= π$ sq units

BITSAT-2015

Conic Section

119625 A point diametrically opposite to the point $P (1$ , 0) on the circle $x^{2} + y^{2} + 2 x + 4 y - 3 = 0$ is

1

(3, - 4)

2

(- 3, 4)

3

(- 3, - 4)

4

(3, 4)

Explanation:

C Given,
Equation circle is $x^{2} + y^{2} + 2 x + 4 y - 3 = 0$
rewritten as, $(x + 1)^{2} + (y + 2)^{2} = (2 \sqrt{2})^{2}$
Let, required point be $Q (α, β)$
Then, mid-point of $P (1, 0)$ and $Q (α, β)$ is the center of the circle $(- 1, - 2)$
Then, $\frac{α + 1}{2} = - 1$
$α + 1 = - 2$
$α = - 3$
and, $\frac{β + 0}{2} = - 2$
$β = - 4$ So, required point is $(- 3, - 4)$

JEEE-2015

Conic Section

119626 The circle $x^{2} + y^{2} + 4 x - 7 y + 12 = 0$ cuts an intercept on y-axis of length

1 3

2 4

3 7

4 1

Explanation:

D circle $x^{2} + y^{2} + 2 gx + 2 fy + c = 0 intercept$
$with y -axis is 2 \sqrt{f^{2} - c}$
$now circle is x^{2} + y^{2} + 4 x - 7 y + 12 = 0 \dots . . (i)$
$Comparing it with equation of circle, we get -$
$f = - \frac{7}{2}, c = 12$
$So, intercept on y-axis of length$
$∴ 2 \sqrt{f^{2} - c} = 2 \sqrt{{(\frac{- 7}{2})}^{2} - 12}$
$= 2 \sqrt{\frac{49}{4} - 12} = 2 \sqrt{\frac{49 - 48}{4}} = \frac{2}{2} = 1$
Ans: d
Exp:D circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ intercept with y-axis is $2 \sqrt{f^{2} - c}$
Comparing it with equation of circle, we get -
So, intercept on y-axis of length

JEEE-2011

Conic Section

119646 The number of integral values of $λ$ for which the equation $x^{2} + y^{2} - 2 λ x + 2 λ y + 14 = 0$ represents a circle whose radius cannot exceed 6 is

1 10

2 11

3 12

4 9

Explanation:

B $x^{2} + y^{2} - 2 λ x + 2 λ y + 14 = 0$
radius $= \sqrt{λ^{2} + λ^{2} - 14}$
We have radius $\leq 6$ .
(radius) $^{2} \leq 36$
$2 λ^{2} - 14 \leq 36$
$2 λ^{2} \leq 50$
$λ^{2} \leq 25$
$λ^{2} - 25 \leq 0$
$- 5 \leq λ \leq 5$
So, number of integral values of $λ$ is 11 .
$λ = - 5, - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, 5$

BCECE-2017

Conic Section

119630 Area of the circle in which a chord of length $\sqrt{2}$ makes an angle $π / 2$ at the centre, is

1

π / 2

squnits

2

2 π

sq units

3

π

sq units

4

π / 4

squnits

Explanation:

C :
original image
Given, chord $(A B) = \sqrt{2}$
$θ = \frac{π}{2}$
Radius of circle $(r) = \frac{A B / 2}{\sin \frac{θ}{2}} = \frac{\sqrt{2} / 2}{\sin \frac{π / 2}{2}}$
$r = \frac{1 / \sqrt{2}}{1 / \sqrt{2}} = 1 unit$
Area of the circle $= π r^{2}$
Area of the circle $= π \times 1^{2}$
Area of the circle $= π$ sq units

BITSAT-2015

Conic Section

119625 A point diametrically opposite to the point $P (1$ , 0) on the circle $x^{2} + y^{2} + 2 x + 4 y - 3 = 0$ is

1

(3, - 4)

2

(- 3, 4)

3

(- 3, - 4)

4

(3, 4)

Explanation:

C Given,
Equation circle is $x^{2} + y^{2} + 2 x + 4 y - 3 = 0$
rewritten as, $(x + 1)^{2} + (y + 2)^{2} = (2 \sqrt{2})^{2}$
Let, required point be $Q (α, β)$
Then, mid-point of $P (1, 0)$ and $Q (α, β)$ is the center of the circle $(- 1, - 2)$
Then, $\frac{α + 1}{2} = - 1$
$α + 1 = - 2$
$α = - 3$
and, $\frac{β + 0}{2} = - 2$
$β = - 4$ So, required point is $(- 3, - 4)$

JEEE-2015

Conic Section

119626 The circle $x^{2} + y^{2} + 4 x - 7 y + 12 = 0$ cuts an intercept on y-axis of length

1 3

2 4

3 7

4 1

Explanation:

D circle $x^{2} + y^{2} + 2 gx + 2 fy + c = 0 intercept$
$with y -axis is 2 \sqrt{f^{2} - c}$
$now circle is x^{2} + y^{2} + 4 x - 7 y + 12 = 0 \dots . . (i)$
$Comparing it with equation of circle, we get -$
$f = - \frac{7}{2}, c = 12$
$So, intercept on y-axis of length$
$∴ 2 \sqrt{f^{2} - c} = 2 \sqrt{{(\frac{- 7}{2})}^{2} - 12}$
$= 2 \sqrt{\frac{49}{4} - 12} = 2 \sqrt{\frac{49 - 48}{4}} = \frac{2}{2} = 1$
Ans: d
Exp:D circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ intercept with y-axis is $2 \sqrt{f^{2} - c}$
Comparing it with equation of circle, we get -
So, intercept on y-axis of length

JEEE-2011

Conic Section

119646 The number of integral values of $λ$ for which the equation $x^{2} + y^{2} - 2 λ x + 2 λ y + 14 = 0$ represents a circle whose radius cannot exceed 6 is

1 10

2 11

3 12

4 9

Explanation:

B $x^{2} + y^{2} - 2 λ x + 2 λ y + 14 = 0$
radius $= \sqrt{λ^{2} + λ^{2} - 14}$
We have radius $\leq 6$ .
(radius) $^{2} \leq 36$
$2 λ^{2} - 14 \leq 36$
$2 λ^{2} \leq 50$
$λ^{2} \leq 25$
$λ^{2} - 25 \leq 0$
$- 5 \leq λ \leq 5$
So, number of integral values of $λ$ is 11 .
$λ = - 5, - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, 5$

BCECE-2017

Conic Section

119630 Area of the circle in which a chord of length $\sqrt{2}$ makes an angle $π / 2$ at the centre, is

1

π / 2

squnits

2

2 π

sq units

3

π

sq units

4

π / 4

squnits

Explanation:

C :
original image
Given, chord $(A B) = \sqrt{2}$
$θ = \frac{π}{2}$
Radius of circle $(r) = \frac{A B / 2}{\sin \frac{θ}{2}} = \frac{\sqrt{2} / 2}{\sin \frac{π / 2}{2}}$
$r = \frac{1 / \sqrt{2}}{1 / \sqrt{2}} = 1 unit$
Area of the circle $= π r^{2}$
Area of the circle $= π \times 1^{2}$
Area of the circle $= π$ sq units

BITSAT-2015

Conic Section

119625 A point diametrically opposite to the point $P (1$ , 0) on the circle $x^{2} + y^{2} + 2 x + 4 y - 3 = 0$ is

1

(3, - 4)

2

(- 3, 4)

3

(- 3, - 4)

4

(3, 4)

Explanation:

C Given,
Equation circle is $x^{2} + y^{2} + 2 x + 4 y - 3 = 0$
rewritten as, $(x + 1)^{2} + (y + 2)^{2} = (2 \sqrt{2})^{2}$
Let, required point be $Q (α, β)$
Then, mid-point of $P (1, 0)$ and $Q (α, β)$ is the center of the circle $(- 1, - 2)$
Then, $\frac{α + 1}{2} = - 1$
$α + 1 = - 2$
$α = - 3$
and, $\frac{β + 0}{2} = - 2$
$β = - 4$ So, required point is $(- 3, - 4)$

JEEE-2015

Conic Section

119626 The circle $x^{2} + y^{2} + 4 x - 7 y + 12 = 0$ cuts an intercept on y-axis of length

1 3

2 4

3 7

4 1

Explanation:

D circle $x^{2} + y^{2} + 2 gx + 2 fy + c = 0 intercept$
$with y -axis is 2 \sqrt{f^{2} - c}$
$now circle is x^{2} + y^{2} + 4 x - 7 y + 12 = 0 \dots . . (i)$
$Comparing it with equation of circle, we get -$
$f = - \frac{7}{2}, c = 12$
$So, intercept on y-axis of length$
$∴ 2 \sqrt{f^{2} - c} = 2 \sqrt{{(\frac{- 7}{2})}^{2} - 12}$
$= 2 \sqrt{\frac{49}{4} - 12} = 2 \sqrt{\frac{49 - 48}{4}} = \frac{2}{2} = 1$
Ans: d
Exp:D circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ intercept with y-axis is $2 \sqrt{f^{2} - c}$
Comparing it with equation of circle, we get -
So, intercept on y-axis of length

JEEE-2011

Conic Section

119646 The number of integral values of $λ$ for which the equation $x^{2} + y^{2} - 2 λ x + 2 λ y + 14 = 0$ represents a circle whose radius cannot exceed 6 is

1 10

2 11

3 12

4 9

Explanation:

B $x^{2} + y^{2} - 2 λ x + 2 λ y + 14 = 0$
radius $= \sqrt{λ^{2} + λ^{2} - 14}$
We have radius $\leq 6$ .
(radius) $^{2} \leq 36$
$2 λ^{2} - 14 \leq 36$
$2 λ^{2} \leq 50$
$λ^{2} \leq 25$
$λ^{2} - 25 \leq 0$
$- 5 \leq λ \leq 5$
So, number of integral values of $λ$ is 11 .
$λ = - 5, - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, 5$

BCECE-2017

Conic Section

119630 Area of the circle in which a chord of length $\sqrt{2}$ makes an angle $π / 2$ at the centre, is

1

π / 2

squnits

2

2 π

sq units

3

π

sq units

4

π / 4

squnits

Explanation:

C :
original image
Given, chord $(A B) = \sqrt{2}$
$θ = \frac{π}{2}$
Radius of circle $(r) = \frac{A B / 2}{\sin \frac{θ}{2}} = \frac{\sqrt{2} / 2}{\sin \frac{π / 2}{2}}$
$r = \frac{1 / \sqrt{2}}{1 / \sqrt{2}} = 1 unit$
Area of the circle $= π r^{2}$
Area of the circle $= π \times 1^{2}$
Area of the circle $= π$ sq units

BITSAT-2015

Conic Section

119625 A point diametrically opposite to the point $P (1$ , 0) on the circle $x^{2} + y^{2} + 2 x + 4 y - 3 = 0$ is

1

(3, - 4)

2

(- 3, 4)

3

(- 3, - 4)

4

(3, 4)

Explanation:

C Given,
Equation circle is $x^{2} + y^{2} + 2 x + 4 y - 3 = 0$
rewritten as, $(x + 1)^{2} + (y + 2)^{2} = (2 \sqrt{2})^{2}$
Let, required point be $Q (α, β)$
Then, mid-point of $P (1, 0)$ and $Q (α, β)$ is the center of the circle $(- 1, - 2)$
Then, $\frac{α + 1}{2} = - 1$
$α + 1 = - 2$
$α = - 3$
and, $\frac{β + 0}{2} = - 2$
$β = - 4$ So, required point is $(- 3, - 4)$

JEEE-2015

Conic Section

119626 The circle $x^{2} + y^{2} + 4 x - 7 y + 12 = 0$ cuts an intercept on y-axis of length

1 3

2 4

3 7

4 1

Explanation:

D circle $x^{2} + y^{2} + 2 gx + 2 fy + c = 0 intercept$
$with y -axis is 2 \sqrt{f^{2} - c}$
$now circle is x^{2} + y^{2} + 4 x - 7 y + 12 = 0 \dots . . (i)$
$Comparing it with equation of circle, we get -$
$f = - \frac{7}{2}, c = 12$
$So, intercept on y-axis of length$
$∴ 2 \sqrt{f^{2} - c} = 2 \sqrt{{(\frac{- 7}{2})}^{2} - 12}$
$= 2 \sqrt{\frac{49}{4} - 12} = 2 \sqrt{\frac{49 - 48}{4}} = \frac{2}{2} = 1$
Ans: d
Exp:D circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ intercept with y-axis is $2 \sqrt{f^{2} - c}$
Comparing it with equation of circle, we get -
So, intercept on y-axis of length

JEEE-2011

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