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Conic Section

119709 If $x^{2} + y^{2} - a^{2} + λ (x \cos α + y \sin α - p) = 0$ is the smallest circle through the points of intersection of $x^{2} + y^{2} = a^{2}$ and $x \cos α + y \sin α =$ $p, 0 < p < a$ , then $λ =$

1 1

2

- p

3

- 2 p

4

- 3 p

Explanation:

C Equation of circle $x^{2} + y^{2} - a^{2} + λ (x \cos α + y \sin α - p) = 0$ is the smallest circle, then centre $(\frac{- λ \cos α}{2}, \frac{- λ \sin α}{2})$ lies on the line
$x \cos α + y \sin α = p$
$∴$ Put centre in the line (i), then we get-
$\frac{- λ \cos^{2} α}{2} - \frac{λ \sin^{2} α}{2} = p$
$\Rightarrow \frac{- λ}{2} = p$
$\Rightarrow λ = - 2 p$

TS EAMCET-11.09.2020

Conic Section

119710 If the poles of the line $x - y = 0$ with respect the circles $x^{2} + y^{2} - 2 g_{i} x + c_{i}^{2} = 0 (i = 1, 2, 3)$ are $(α_{i}$ , $β_{i})$ , then $\sum_{i = 1}^{3} \frac{α_{i} + β_{i}}{g_{i}} =$

1 3

2 6

3

\frac{3}{2}

4

\frac{3}{4}

Explanation:

A Since equation polar of point $(α_{i}, β_{i})$ with respect to the circle $x^{2} + y^{2} - 2 g_{i} x + c_{i}^{2} = 0$ is
$α_{i} x + β_{i} y - g_{i} (x + α_{i}) + c_{i}^{2} = 0$
$(α_{i} - g_{i}) x + β_{i} y + c_{i}^{2} - α_{i} g_{i} = 0$
Now, on comparing with line $x - y = 0$ , we get
$\frac{α_{i} - g_{i}}{1} = \frac{β_{i}}{- 1} = \frac{c_{i}^{2} - α_{i} g_{i}}{0}$
$c_{i}^{2} = α_{i} g_{i} and α_{i} + β_{i} = g_{i}$
$∴ \sum_{i = 1}^{3} \frac{α_{i} + β_{i}}{g_{i}} = \sum_{i = 1}^{3} (1) = 3$

TS EAMCET-11.09.2020

Conic Section

119711 If the circle $x^{2} + y^{2} = a^{2}$ intersects the hyperbola $x y = b^{2}$ at four points $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ , $(x_{4}, y_{4})$ , then $y_{1} y_{2} y_{3} y_{4} =$

1 a

2 0

3

b^{4}

4

b^{2}

Explanation:

C We have,
And
$x^{2} + y^{2} = a^{2}$
$xy = b^{2}$
From Eq. (ii) $x = \frac{b^{2}}{y}$
From Eq. (i) ${(\frac{b^{2}}{y})}^{2} + y^{2} = a^{2}$
$b^{4} + y^{4} = a^{2} y^{2}$
$y^{4} - a^{2} y^{2} + b^{4} = 0$ This is an equation of 4th degree in $y$ and its four roots are $y_{1}, y_{2}, y_{3}, y_{4}$ then $y_{1} y_{2} y_{3} y_{4} = b^{4}$

TS EAMCET-10.09.2020

Conic Section

119712 If $α \neq - 4$ and $(2, α)$ is the mid-point of a chord of the circle $x^{2} + y^{2} - 4 x + 8 y + 6 = 0$ , then the values of the $y$ -intercept of the chord lie in the interval

1

(- 4 - \sqrt{14}, - 4 + \sqrt{14})

2

(- 4, 4)

3

(4, \sqrt{14}, 4 + \sqrt{14})

4

(- 2, 2)

Explanation:

A We have,
$(2, α)$ is mid-point of chord of circle
$x^{2} + y^{2} - 4 x + 8 y + 6 = 0$
$∵ (2, α)$ inside the circle
$∴ 4 + α^{2} - 8 + 8 α + 6 \leq 0$
$α^{2} + 8 α + 2 \leq 0$
$α = \frac{- 8 \pm \sqrt{64 - 8}}{2}$
$α = \frac{- 8 \pm 2 \sqrt{14}}{2} = - 4 \pm \sqrt{14}$
$α \in (- 4 - \sqrt{14}, - 4 + \sqrt{14})$

TS EAMCET-10.09.2020

Conic Section

119709 If $x^{2} + y^{2} - a^{2} + λ (x \cos α + y \sin α - p) = 0$ is the smallest circle through the points of intersection of $x^{2} + y^{2} = a^{2}$ and $x \cos α + y \sin α =$ $p, 0 < p < a$ , then $λ =$

1 1

2

- p

3

- 2 p

4

- 3 p

Explanation:

C Equation of circle $x^{2} + y^{2} - a^{2} + λ (x \cos α + y \sin α - p) = 0$ is the smallest circle, then centre $(\frac{- λ \cos α}{2}, \frac{- λ \sin α}{2})$ lies on the line
$x \cos α + y \sin α = p$
$∴$ Put centre in the line (i), then we get-
$\frac{- λ \cos^{2} α}{2} - \frac{λ \sin^{2} α}{2} = p$
$\Rightarrow \frac{- λ}{2} = p$
$\Rightarrow λ = - 2 p$

TS EAMCET-11.09.2020

Conic Section

119710 If the poles of the line $x - y = 0$ with respect the circles $x^{2} + y^{2} - 2 g_{i} x + c_{i}^{2} = 0 (i = 1, 2, 3)$ are $(α_{i}$ , $β_{i})$ , then $\sum_{i = 1}^{3} \frac{α_{i} + β_{i}}{g_{i}} =$

1 3

2 6

3

\frac{3}{2}

4

\frac{3}{4}

Explanation:

A Since equation polar of point $(α_{i}, β_{i})$ with respect to the circle $x^{2} + y^{2} - 2 g_{i} x + c_{i}^{2} = 0$ is
$α_{i} x + β_{i} y - g_{i} (x + α_{i}) + c_{i}^{2} = 0$
$(α_{i} - g_{i}) x + β_{i} y + c_{i}^{2} - α_{i} g_{i} = 0$
Now, on comparing with line $x - y = 0$ , we get
$\frac{α_{i} - g_{i}}{1} = \frac{β_{i}}{- 1} = \frac{c_{i}^{2} - α_{i} g_{i}}{0}$
$c_{i}^{2} = α_{i} g_{i} and α_{i} + β_{i} = g_{i}$
$∴ \sum_{i = 1}^{3} \frac{α_{i} + β_{i}}{g_{i}} = \sum_{i = 1}^{3} (1) = 3$

TS EAMCET-11.09.2020

Conic Section

119711 If the circle $x^{2} + y^{2} = a^{2}$ intersects the hyperbola $x y = b^{2}$ at four points $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ , $(x_{4}, y_{4})$ , then $y_{1} y_{2} y_{3} y_{4} =$

1 a

2 0

3

b^{4}

4

b^{2}

Explanation:

C We have,
And
$x^{2} + y^{2} = a^{2}$
$xy = b^{2}$
From Eq. (ii) $x = \frac{b^{2}}{y}$
From Eq. (i) ${(\frac{b^{2}}{y})}^{2} + y^{2} = a^{2}$
$b^{4} + y^{4} = a^{2} y^{2}$
$y^{4} - a^{2} y^{2} + b^{4} = 0$ This is an equation of 4th degree in $y$ and its four roots are $y_{1}, y_{2}, y_{3}, y_{4}$ then $y_{1} y_{2} y_{3} y_{4} = b^{4}$

TS EAMCET-10.09.2020

Conic Section

119712 If $α \neq - 4$ and $(2, α)$ is the mid-point of a chord of the circle $x^{2} + y^{2} - 4 x + 8 y + 6 = 0$ , then the values of the $y$ -intercept of the chord lie in the interval

1

(- 4 - \sqrt{14}, - 4 + \sqrt{14})

2

(- 4, 4)

3

(4, \sqrt{14}, 4 + \sqrt{14})

4

(- 2, 2)

Explanation:

A We have,
$(2, α)$ is mid-point of chord of circle
$x^{2} + y^{2} - 4 x + 8 y + 6 = 0$
$∵ (2, α)$ inside the circle
$∴ 4 + α^{2} - 8 + 8 α + 6 \leq 0$
$α^{2} + 8 α + 2 \leq 0$
$α = \frac{- 8 \pm \sqrt{64 - 8}}{2}$
$α = \frac{- 8 \pm 2 \sqrt{14}}{2} = - 4 \pm \sqrt{14}$
$α \in (- 4 - \sqrt{14}, - 4 + \sqrt{14})$

TS EAMCET-10.09.2020

Conic Section

119709 If $x^{2} + y^{2} - a^{2} + λ (x \cos α + y \sin α - p) = 0$ is the smallest circle through the points of intersection of $x^{2} + y^{2} = a^{2}$ and $x \cos α + y \sin α =$ $p, 0 < p < a$ , then $λ =$

1 1

2

- p

3

- 2 p

4

- 3 p

Explanation:

C Equation of circle $x^{2} + y^{2} - a^{2} + λ (x \cos α + y \sin α - p) = 0$ is the smallest circle, then centre $(\frac{- λ \cos α}{2}, \frac{- λ \sin α}{2})$ lies on the line
$x \cos α + y \sin α = p$
$∴$ Put centre in the line (i), then we get-
$\frac{- λ \cos^{2} α}{2} - \frac{λ \sin^{2} α}{2} = p$
$\Rightarrow \frac{- λ}{2} = p$
$\Rightarrow λ = - 2 p$

TS EAMCET-11.09.2020

Conic Section

119710 If the poles of the line $x - y = 0$ with respect the circles $x^{2} + y^{2} - 2 g_{i} x + c_{i}^{2} = 0 (i = 1, 2, 3)$ are $(α_{i}$ , $β_{i})$ , then $\sum_{i = 1}^{3} \frac{α_{i} + β_{i}}{g_{i}} =$

1 3

2 6

3

\frac{3}{2}

4

\frac{3}{4}

Explanation:

A Since equation polar of point $(α_{i}, β_{i})$ with respect to the circle $x^{2} + y^{2} - 2 g_{i} x + c_{i}^{2} = 0$ is
$α_{i} x + β_{i} y - g_{i} (x + α_{i}) + c_{i}^{2} = 0$
$(α_{i} - g_{i}) x + β_{i} y + c_{i}^{2} - α_{i} g_{i} = 0$
Now, on comparing with line $x - y = 0$ , we get
$\frac{α_{i} - g_{i}}{1} = \frac{β_{i}}{- 1} = \frac{c_{i}^{2} - α_{i} g_{i}}{0}$
$c_{i}^{2} = α_{i} g_{i} and α_{i} + β_{i} = g_{i}$
$∴ \sum_{i = 1}^{3} \frac{α_{i} + β_{i}}{g_{i}} = \sum_{i = 1}^{3} (1) = 3$

TS EAMCET-11.09.2020

Conic Section

119711 If the circle $x^{2} + y^{2} = a^{2}$ intersects the hyperbola $x y = b^{2}$ at four points $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ , $(x_{4}, y_{4})$ , then $y_{1} y_{2} y_{3} y_{4} =$

1 a

2 0

3

b^{4}

4

b^{2}

Explanation:

C We have,
And
$x^{2} + y^{2} = a^{2}$
$xy = b^{2}$
From Eq. (ii) $x = \frac{b^{2}}{y}$
From Eq. (i) ${(\frac{b^{2}}{y})}^{2} + y^{2} = a^{2}$
$b^{4} + y^{4} = a^{2} y^{2}$
$y^{4} - a^{2} y^{2} + b^{4} = 0$ This is an equation of 4th degree in $y$ and its four roots are $y_{1}, y_{2}, y_{3}, y_{4}$ then $y_{1} y_{2} y_{3} y_{4} = b^{4}$

TS EAMCET-10.09.2020

Conic Section

119712 If $α \neq - 4$ and $(2, α)$ is the mid-point of a chord of the circle $x^{2} + y^{2} - 4 x + 8 y + 6 = 0$ , then the values of the $y$ -intercept of the chord lie in the interval

1

(- 4 - \sqrt{14}, - 4 + \sqrt{14})

2

(- 4, 4)

3

(4, \sqrt{14}, 4 + \sqrt{14})

4

(- 2, 2)

Explanation:

A We have,
$(2, α)$ is mid-point of chord of circle
$x^{2} + y^{2} - 4 x + 8 y + 6 = 0$
$∵ (2, α)$ inside the circle
$∴ 4 + α^{2} - 8 + 8 α + 6 \leq 0$
$α^{2} + 8 α + 2 \leq 0$
$α = \frac{- 8 \pm \sqrt{64 - 8}}{2}$
$α = \frac{- 8 \pm 2 \sqrt{14}}{2} = - 4 \pm \sqrt{14}$
$α \in (- 4 - \sqrt{14}, - 4 + \sqrt{14})$

TS EAMCET-10.09.2020

Conic Section

119709 If $x^{2} + y^{2} - a^{2} + λ (x \cos α + y \sin α - p) = 0$ is the smallest circle through the points of intersection of $x^{2} + y^{2} = a^{2}$ and $x \cos α + y \sin α =$ $p, 0 < p < a$ , then $λ =$

1 1

2

- p

3

- 2 p

4

- 3 p

Explanation:

C Equation of circle $x^{2} + y^{2} - a^{2} + λ (x \cos α + y \sin α - p) = 0$ is the smallest circle, then centre $(\frac{- λ \cos α}{2}, \frac{- λ \sin α}{2})$ lies on the line
$x \cos α + y \sin α = p$
$∴$ Put centre in the line (i), then we get-
$\frac{- λ \cos^{2} α}{2} - \frac{λ \sin^{2} α}{2} = p$
$\Rightarrow \frac{- λ}{2} = p$
$\Rightarrow λ = - 2 p$

TS EAMCET-11.09.2020

Conic Section

119710 If the poles of the line $x - y = 0$ with respect the circles $x^{2} + y^{2} - 2 g_{i} x + c_{i}^{2} = 0 (i = 1, 2, 3)$ are $(α_{i}$ , $β_{i})$ , then $\sum_{i = 1}^{3} \frac{α_{i} + β_{i}}{g_{i}} =$

1 3

2 6

3

\frac{3}{2}

4

\frac{3}{4}

Explanation:

A Since equation polar of point $(α_{i}, β_{i})$ with respect to the circle $x^{2} + y^{2} - 2 g_{i} x + c_{i}^{2} = 0$ is
$α_{i} x + β_{i} y - g_{i} (x + α_{i}) + c_{i}^{2} = 0$
$(α_{i} - g_{i}) x + β_{i} y + c_{i}^{2} - α_{i} g_{i} = 0$
Now, on comparing with line $x - y = 0$ , we get
$\frac{α_{i} - g_{i}}{1} = \frac{β_{i}}{- 1} = \frac{c_{i}^{2} - α_{i} g_{i}}{0}$
$c_{i}^{2} = α_{i} g_{i} and α_{i} + β_{i} = g_{i}$
$∴ \sum_{i = 1}^{3} \frac{α_{i} + β_{i}}{g_{i}} = \sum_{i = 1}^{3} (1) = 3$

TS EAMCET-11.09.2020

Conic Section

119711 If the circle $x^{2} + y^{2} = a^{2}$ intersects the hyperbola $x y = b^{2}$ at four points $(x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})$ , $(x_{4}, y_{4})$ , then $y_{1} y_{2} y_{3} y_{4} =$

1 a

2 0

3

b^{4}

4

b^{2}

Explanation:

C We have,
And
$x^{2} + y^{2} = a^{2}$
$xy = b^{2}$
From Eq. (ii) $x = \frac{b^{2}}{y}$
From Eq. (i) ${(\frac{b^{2}}{y})}^{2} + y^{2} = a^{2}$
$b^{4} + y^{4} = a^{2} y^{2}$
$y^{4} - a^{2} y^{2} + b^{4} = 0$ This is an equation of 4th degree in $y$ and its four roots are $y_{1}, y_{2}, y_{3}, y_{4}$ then $y_{1} y_{2} y_{3} y_{4} = b^{4}$

TS EAMCET-10.09.2020

Conic Section

119712 If $α \neq - 4$ and $(2, α)$ is the mid-point of a chord of the circle $x^{2} + y^{2} - 4 x + 8 y + 6 = 0$ , then the values of the $y$ -intercept of the chord lie in the interval

1

(- 4 - \sqrt{14}, - 4 + \sqrt{14})

2

(- 4, 4)

3

(4, \sqrt{14}, 4 + \sqrt{14})

4

(- 2, 2)

Explanation:

A We have,
$(2, α)$ is mid-point of chord of circle
$x^{2} + y^{2} - 4 x + 8 y + 6 = 0$
$∵ (2, α)$ inside the circle
$∴ 4 + α^{2} - 8 + 8 α + 6 \leq 0$
$α^{2} + 8 α + 2 \leq 0$
$α = \frac{- 8 \pm \sqrt{64 - 8}}{2}$
$α = \frac{- 8 \pm 2 \sqrt{14}}{2} = - 4 \pm \sqrt{14}$
$α \in (- 4 - \sqrt{14}, - 4 + \sqrt{14})$

TS EAMCET-10.09.2020