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Conic Section

119695 The radius of the circle
$2 x^{2} + 2 y^{2} - 3 x + 2 y - 1 = 0 \dots . .$ . units.

1

\frac{\sqrt{21}}{2}

2

\frac{\sqrt{21}}{4}

3

\frac{21}{4}

4

\frac{\sqrt{5}}{4}

Explanation:

B Given circle equation
$2 x^{2} + 2 y^{2} - 3 x + 2 y - 1 = 0$
$x^{2} + y^{2} - \frac{3}{2} x + y - \frac{1}{2} = 0$
By comparing above equation with general equation of circle
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
$g = \frac{- 3}{4}, f = \frac{1}{2}, c = - \frac{1}{2}$
Radius $(R) = \sqrt{(g)^{2} + (f)^{2} - c}$
$= \sqrt{{(\frac{- 3}{4})}^{2} + {(\frac{1}{2})}^{2} - (- \frac{1}{2})}$
$= \sqrt{\frac{9}{16} + \frac{1}{4} + \frac{1}{2}} = \sqrt{\frac{21}{16}}$
$R = \frac{\sqrt{21}}{4}$

AP EAMCET-23.09.2020

Conic Section

119696 A circle having centre at the origin passes through the three vertices of an equilateral triangle the length of its median being 9 units. Then the equation of that circle is

1

x^{2} + y^{2} = 9

2

x^{2} + y^{2} = 18

3

x^{2} + y^{2} = 36

4

x^{2} + y^{2} = 81

Explanation:

C Let the equation of a circle passing through one vertex of an equilateral triangle whose median of length $3 a$ and having centre at origin is-
$x^{2} + y^{2} = 4 a^{2}$
Given that, the length of the median is 9 units
Hence, $3 a = 9$
$a = 3$
Therefore the equation of the circle is-
$x^{2} + y^{2} = 4 a^{2}$
$x^{2} + y^{2} = 4 \times (3)^{2}$
$x^{2} + y^{2} = 36$

AP EAMCET-2017

Conic Section

119697 The point to which the origin is to be shifted to remove the first degree terms from the equation $2 x^{2} + 4 x y - 6 y^{2} + 2 x + 8 y + 1 = 0$ is

1

(\frac{7}{8}, \frac{3}{8})

2

(\frac{- 7}{8}, \frac{- 3}{8})

3

(\frac{- 7}{8}, \frac{3}{8})

4

(\frac{7}{8}, \frac{- 3}{8})

Explanation:

C Given equation -
$F = 2 x^{2} + 4 x y - 6 y^{2} + 2 x + 8 y + 1 = 0$
On partial differential we get
$\frac{\partial F}{\partial x} = 4 x + 4 y + 2 = 0$
$\frac{\partial F}{\partial y} = - 12 y + 4 x + 8 = 0$
From solving equation (i) \& (ii), we get
$4 x + 4 y + 2 = 0$
$4 x - 12 y + 8 = 0$
$+ -$
$16 y - 6 = 0$
$y = \frac{+ 6}{16} = \frac{3}{8}$
By putting the value of $y$ in equation (i) we get
$4 x + 4 \times \frac{3}{8} + 2 = 0$
$4 x = - \frac{7}{2} \Rightarrow x = - \frac{7}{8}$ Hence $(- \frac{7}{8}, \frac{3}{8})$

AP EAMCET-2017

Conic Section

119698 If a circle with radius 2.5 units passes through the points $(2, 3)$ and $(5, 7)$ , then its centre is

1

(1.5, 2)

2

(7, 10)

3

(3, 4)

4

(3.5, 5)

Explanation:

D $CA = CB$
${CA}^{2} = {CB}^{2}$
original image
$d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$
$C A = \sqrt{(x - 2)^{2} + (y - 3)^{2}} = \sqrt{(x - 5)^{2} + (y - 2)^{2}} = C B$
$C^{2} = C^{2}$
$(x - 2)^{2} + (y - 3)^{2} = (x - 5)^{2} + (y - 7)^{2}$
$x^{2} + 4 - 4 x + y^{2} + 9 - 6 y = x^{2} + 25 - 10 x + y^{2} + 49 - 14 y$
$6 x + 8 y = 61$ The point $(3.5, 5)$ only satisfy the above equation (i) thus $(x, y) = (3.5, 5)$

AP EAMCET-2017

Conic Section

119695 The radius of the circle
$2 x^{2} + 2 y^{2} - 3 x + 2 y - 1 = 0 \dots . .$ . units.

1

\frac{\sqrt{21}}{2}

2

\frac{\sqrt{21}}{4}

3

\frac{21}{4}

4

\frac{\sqrt{5}}{4}

Explanation:

B Given circle equation
$2 x^{2} + 2 y^{2} - 3 x + 2 y - 1 = 0$
$x^{2} + y^{2} - \frac{3}{2} x + y - \frac{1}{2} = 0$
By comparing above equation with general equation of circle
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
$g = \frac{- 3}{4}, f = \frac{1}{2}, c = - \frac{1}{2}$
Radius $(R) = \sqrt{(g)^{2} + (f)^{2} - c}$
$= \sqrt{{(\frac{- 3}{4})}^{2} + {(\frac{1}{2})}^{2} - (- \frac{1}{2})}$
$= \sqrt{\frac{9}{16} + \frac{1}{4} + \frac{1}{2}} = \sqrt{\frac{21}{16}}$
$R = \frac{\sqrt{21}}{4}$

AP EAMCET-23.09.2020

Conic Section

119696 A circle having centre at the origin passes through the three vertices of an equilateral triangle the length of its median being 9 units. Then the equation of that circle is

1

x^{2} + y^{2} = 9

2

x^{2} + y^{2} = 18

3

x^{2} + y^{2} = 36

4

x^{2} + y^{2} = 81

Explanation:

C Let the equation of a circle passing through one vertex of an equilateral triangle whose median of length $3 a$ and having centre at origin is-
$x^{2} + y^{2} = 4 a^{2}$
Given that, the length of the median is 9 units
Hence, $3 a = 9$
$a = 3$
Therefore the equation of the circle is-
$x^{2} + y^{2} = 4 a^{2}$
$x^{2} + y^{2} = 4 \times (3)^{2}$
$x^{2} + y^{2} = 36$

AP EAMCET-2017

Conic Section

119697 The point to which the origin is to be shifted to remove the first degree terms from the equation $2 x^{2} + 4 x y - 6 y^{2} + 2 x + 8 y + 1 = 0$ is

1

(\frac{7}{8}, \frac{3}{8})

2

(\frac{- 7}{8}, \frac{- 3}{8})

3

(\frac{- 7}{8}, \frac{3}{8})

4

(\frac{7}{8}, \frac{- 3}{8})

Explanation:

C Given equation -
$F = 2 x^{2} + 4 x y - 6 y^{2} + 2 x + 8 y + 1 = 0$
On partial differential we get
$\frac{\partial F}{\partial x} = 4 x + 4 y + 2 = 0$
$\frac{\partial F}{\partial y} = - 12 y + 4 x + 8 = 0$
From solving equation (i) \& (ii), we get
$4 x + 4 y + 2 = 0$
$4 x - 12 y + 8 = 0$
$+ -$
$16 y - 6 = 0$
$y = \frac{+ 6}{16} = \frac{3}{8}$
By putting the value of $y$ in equation (i) we get
$4 x + 4 \times \frac{3}{8} + 2 = 0$
$4 x = - \frac{7}{2} \Rightarrow x = - \frac{7}{8}$ Hence $(- \frac{7}{8}, \frac{3}{8})$

AP EAMCET-2017

Conic Section

119698 If a circle with radius 2.5 units passes through the points $(2, 3)$ and $(5, 7)$ , then its centre is

1

(1.5, 2)

2

(7, 10)

3

(3, 4)

4

(3.5, 5)

Explanation:

D $CA = CB$
${CA}^{2} = {CB}^{2}$
original image
$d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$
$C A = \sqrt{(x - 2)^{2} + (y - 3)^{2}} = \sqrt{(x - 5)^{2} + (y - 2)^{2}} = C B$
$C^{2} = C^{2}$
$(x - 2)^{2} + (y - 3)^{2} = (x - 5)^{2} + (y - 7)^{2}$
$x^{2} + 4 - 4 x + y^{2} + 9 - 6 y = x^{2} + 25 - 10 x + y^{2} + 49 - 14 y$
$6 x + 8 y = 61$ The point $(3.5, 5)$ only satisfy the above equation (i) thus $(x, y) = (3.5, 5)$

AP EAMCET-2017

Conic Section

119695 The radius of the circle
$2 x^{2} + 2 y^{2} - 3 x + 2 y - 1 = 0 \dots . .$ . units.

1

\frac{\sqrt{21}}{2}

2

\frac{\sqrt{21}}{4}

3

\frac{21}{4}

4

\frac{\sqrt{5}}{4}

Explanation:

B Given circle equation
$2 x^{2} + 2 y^{2} - 3 x + 2 y - 1 = 0$
$x^{2} + y^{2} - \frac{3}{2} x + y - \frac{1}{2} = 0$
By comparing above equation with general equation of circle
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
$g = \frac{- 3}{4}, f = \frac{1}{2}, c = - \frac{1}{2}$
Radius $(R) = \sqrt{(g)^{2} + (f)^{2} - c}$
$= \sqrt{{(\frac{- 3}{4})}^{2} + {(\frac{1}{2})}^{2} - (- \frac{1}{2})}$
$= \sqrt{\frac{9}{16} + \frac{1}{4} + \frac{1}{2}} = \sqrt{\frac{21}{16}}$
$R = \frac{\sqrt{21}}{4}$

AP EAMCET-23.09.2020

Conic Section

119696 A circle having centre at the origin passes through the three vertices of an equilateral triangle the length of its median being 9 units. Then the equation of that circle is

1

x^{2} + y^{2} = 9

2

x^{2} + y^{2} = 18

3

x^{2} + y^{2} = 36

4

x^{2} + y^{2} = 81

Explanation:

C Let the equation of a circle passing through one vertex of an equilateral triangle whose median of length $3 a$ and having centre at origin is-
$x^{2} + y^{2} = 4 a^{2}$
Given that, the length of the median is 9 units
Hence, $3 a = 9$
$a = 3$
Therefore the equation of the circle is-
$x^{2} + y^{2} = 4 a^{2}$
$x^{2} + y^{2} = 4 \times (3)^{2}$
$x^{2} + y^{2} = 36$

AP EAMCET-2017

Conic Section

119697 The point to which the origin is to be shifted to remove the first degree terms from the equation $2 x^{2} + 4 x y - 6 y^{2} + 2 x + 8 y + 1 = 0$ is

1

(\frac{7}{8}, \frac{3}{8})

2

(\frac{- 7}{8}, \frac{- 3}{8})

3

(\frac{- 7}{8}, \frac{3}{8})

4

(\frac{7}{8}, \frac{- 3}{8})

Explanation:

C Given equation -
$F = 2 x^{2} + 4 x y - 6 y^{2} + 2 x + 8 y + 1 = 0$
On partial differential we get
$\frac{\partial F}{\partial x} = 4 x + 4 y + 2 = 0$
$\frac{\partial F}{\partial y} = - 12 y + 4 x + 8 = 0$
From solving equation (i) \& (ii), we get
$4 x + 4 y + 2 = 0$
$4 x - 12 y + 8 = 0$
$+ -$
$16 y - 6 = 0$
$y = \frac{+ 6}{16} = \frac{3}{8}$
By putting the value of $y$ in equation (i) we get
$4 x + 4 \times \frac{3}{8} + 2 = 0$
$4 x = - \frac{7}{2} \Rightarrow x = - \frac{7}{8}$ Hence $(- \frac{7}{8}, \frac{3}{8})$

AP EAMCET-2017

Conic Section

119698 If a circle with radius 2.5 units passes through the points $(2, 3)$ and $(5, 7)$ , then its centre is

1

(1.5, 2)

2

(7, 10)

3

(3, 4)

4

(3.5, 5)

Explanation:

D $CA = CB$
${CA}^{2} = {CB}^{2}$
original image
$d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$
$C A = \sqrt{(x - 2)^{2} + (y - 3)^{2}} = \sqrt{(x - 5)^{2} + (y - 2)^{2}} = C B$
$C^{2} = C^{2}$
$(x - 2)^{2} + (y - 3)^{2} = (x - 5)^{2} + (y - 7)^{2}$
$x^{2} + 4 - 4 x + y^{2} + 9 - 6 y = x^{2} + 25 - 10 x + y^{2} + 49 - 14 y$
$6 x + 8 y = 61$ The point $(3.5, 5)$ only satisfy the above equation (i) thus $(x, y) = (3.5, 5)$

AP EAMCET-2017

Conic Section

119695 The radius of the circle
$2 x^{2} + 2 y^{2} - 3 x + 2 y - 1 = 0 \dots . .$ . units.

1

\frac{\sqrt{21}}{2}

2

\frac{\sqrt{21}}{4}

3

\frac{21}{4}

4

\frac{\sqrt{5}}{4}

Explanation:

B Given circle equation
$2 x^{2} + 2 y^{2} - 3 x + 2 y - 1 = 0$
$x^{2} + y^{2} - \frac{3}{2} x + y - \frac{1}{2} = 0$
By comparing above equation with general equation of circle
$x^{2} + y^{2} + 2 gx + 2 fy + c = 0$
$g = \frac{- 3}{4}, f = \frac{1}{2}, c = - \frac{1}{2}$
Radius $(R) = \sqrt{(g)^{2} + (f)^{2} - c}$
$= \sqrt{{(\frac{- 3}{4})}^{2} + {(\frac{1}{2})}^{2} - (- \frac{1}{2})}$
$= \sqrt{\frac{9}{16} + \frac{1}{4} + \frac{1}{2}} = \sqrt{\frac{21}{16}}$
$R = \frac{\sqrt{21}}{4}$

AP EAMCET-23.09.2020

Conic Section

119696 A circle having centre at the origin passes through the three vertices of an equilateral triangle the length of its median being 9 units. Then the equation of that circle is

1

x^{2} + y^{2} = 9

2

x^{2} + y^{2} = 18

3

x^{2} + y^{2} = 36

4

x^{2} + y^{2} = 81

Explanation:

C Let the equation of a circle passing through one vertex of an equilateral triangle whose median of length $3 a$ and having centre at origin is-
$x^{2} + y^{2} = 4 a^{2}$
Given that, the length of the median is 9 units
Hence, $3 a = 9$
$a = 3$
Therefore the equation of the circle is-
$x^{2} + y^{2} = 4 a^{2}$
$x^{2} + y^{2} = 4 \times (3)^{2}$
$x^{2} + y^{2} = 36$

AP EAMCET-2017

Conic Section

119697 The point to which the origin is to be shifted to remove the first degree terms from the equation $2 x^{2} + 4 x y - 6 y^{2} + 2 x + 8 y + 1 = 0$ is

1

(\frac{7}{8}, \frac{3}{8})

2

(\frac{- 7}{8}, \frac{- 3}{8})

3

(\frac{- 7}{8}, \frac{3}{8})

4

(\frac{7}{8}, \frac{- 3}{8})

Explanation:

C Given equation -
$F = 2 x^{2} + 4 x y - 6 y^{2} + 2 x + 8 y + 1 = 0$
On partial differential we get
$\frac{\partial F}{\partial x} = 4 x + 4 y + 2 = 0$
$\frac{\partial F}{\partial y} = - 12 y + 4 x + 8 = 0$
From solving equation (i) \& (ii), we get
$4 x + 4 y + 2 = 0$
$4 x - 12 y + 8 = 0$
$+ -$
$16 y - 6 = 0$
$y = \frac{+ 6}{16} = \frac{3}{8}$
By putting the value of $y$ in equation (i) we get
$4 x + 4 \times \frac{3}{8} + 2 = 0$
$4 x = - \frac{7}{2} \Rightarrow x = - \frac{7}{8}$ Hence $(- \frac{7}{8}, \frac{3}{8})$

AP EAMCET-2017

Conic Section

119698 If a circle with radius 2.5 units passes through the points $(2, 3)$ and $(5, 7)$ , then its centre is

1

(1.5, 2)

2

(7, 10)

3

(3, 4)

4

(3.5, 5)

Explanation:

D $CA = CB$
${CA}^{2} = {CB}^{2}$
original image
$d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$
$C A = \sqrt{(x - 2)^{2} + (y - 3)^{2}} = \sqrt{(x - 5)^{2} + (y - 2)^{2}} = C B$
$C^{2} = C^{2}$
$(x - 2)^{2} + (y - 3)^{2} = (x - 5)^{2} + (y - 7)^{2}$
$x^{2} + 4 - 4 x + y^{2} + 9 - 6 y = x^{2} + 25 - 10 x + y^{2} + 49 - 14 y$
$6 x + 8 y = 61$ The point $(3.5, 5)$ only satisfy the above equation (i) thus $(x, y) = (3.5, 5)$

AP EAMCET-2017