QBManager

Conic Section

119895 The angle between the two tangents drawn from origin to the circle $x^{2} + y^{2} - 14 x + 2 y + 25$ $= 0$

1

0^{\circ}

2

45^{\circ}

3

90^{\circ}

4

60^{\circ}

Explanation:

C : original image
If $2 α$ is the angle between the two tangents from the origin to the circle.
$\tan α = \frac{{CT}_{1}}{{OT}_{1}} = \frac{\sqrt{49 + 1 - 25}}{\sqrt{25}} = 1$
$α = \tan^{- 1} (1) = 45^{\circ}$
$2 α = 2 \times 45^{\circ} = 90^{\circ} = \frac{π}{2}$ Angle between the tangents is $\frac{π}{2}$

AP EAMCET-23.08.2021

Conic Section

119896 The line $a x + b y + c = 0$ is normal to the circle $x^{2} + y^{2} + 2 g x + 2 f y + d = 0$ if

1

a g + b f + c = 0

2

\overset{―}{a g + b} f - c = 0

3

ag - bf + c = 0

4 ag

- bf - c = 0

Explanation:

B The centre of given circle is $(- g, - f)$ . If the
given line $ax + bx + c = 0$ is normal to the circle, then it
passes through the centre of circle
$∴ a (- g) + b (- f) + c = 0$
$ag + bf - c = 0$

AP EAMCET-23.08.2021

Conic Section

119897 The shortest distance from the line $3 x + 4 y = 25$ to the circle $x^{2} + y^{2} - 6 x + 8 y = 0$ is

1

\frac{9}{5}

2

\frac{7}{5}

3

\frac{8}{5}

4

\frac{13}{5}

Explanation:

B Given, equation of circle is
$x^{2} + y^{2} - 6 x + 8 y = 0$
$(x - 3)^{2} + (y + 4)^{2} = 25$
shortest distance $(d) =$ distance of line from centre - Radius
$= \frac{32}{5} - 5 = \frac{32 - 25}{5} d = \frac{7}{5}$

AP EAMCET-08.07.2022

Conic Section

119898 A circle touches the Y-axis at the point $(0, 4)$ and passes through the point $(2, 0)$ . Which of the following lines is not a tangent to this circle?

1

4 x - 3 y + 17 = 0

2

3 x + 4 y - 6 = 0

3

4 x + 3 y - 8 = 0

4

3 x - 4 y - 24 = 0

Explanation:

C Given point $(0, 4)$
Equation of family of circle
$(x - 0)^{2} + (y - 4)^{2} + λ x = 0$ passed through $(2, 0)$
$∴ (2 - 0)^{2} + (0 - 4)^{2} + 2 λ = 0$
$2^{2} + (- 4)^{2} + 2 λ = 0$
$4 + 16 + 2 λ = 0$
$λ = - 10$
Equation of circle
$x^{2} + y^{2} + 16 - 8 y + (- 10 x) = 0$
$x^{2} + y^{2} - 10 x - 8 y + 16 = 0$
centre, $(+ 5, 4)$
Radius $= \sqrt{25 + 16 - 16} = \sqrt{25} = 5$
$\Rightarrow$ checking option $A$
tangent $4 x + 3 y - 8 = 0$ distance from centre $= 5$
$| \frac{(4 \times 5) + (3 \times 4) - 8}{\sqrt{(4)^{2} + (3)^{2}}} | = | \frac{20 + 12 - 8}{5} | = | \frac{24}{5} | \neq 5$
$∴ 4 x + 3 y - 8 = 0$ not a tangent to circle.

JEE Main 09.01.2020

Conic Section

119895 The angle between the two tangents drawn from origin to the circle $x^{2} + y^{2} - 14 x + 2 y + 25$ $= 0$

1

0^{\circ}

2

45^{\circ}

3

90^{\circ}

4

60^{\circ}

Explanation:

C : original image
If $2 α$ is the angle between the two tangents from the origin to the circle.
$\tan α = \frac{{CT}_{1}}{{OT}_{1}} = \frac{\sqrt{49 + 1 - 25}}{\sqrt{25}} = 1$
$α = \tan^{- 1} (1) = 45^{\circ}$
$2 α = 2 \times 45^{\circ} = 90^{\circ} = \frac{π}{2}$ Angle between the tangents is $\frac{π}{2}$

AP EAMCET-23.08.2021

Conic Section

119896 The line $a x + b y + c = 0$ is normal to the circle $x^{2} + y^{2} + 2 g x + 2 f y + d = 0$ if

1

a g + b f + c = 0

2

\overset{―}{a g + b} f - c = 0

3

ag - bf + c = 0

4 ag

- bf - c = 0

Explanation:

B The centre of given circle is $(- g, - f)$ . If the
given line $ax + bx + c = 0$ is normal to the circle, then it
passes through the centre of circle
$∴ a (- g) + b (- f) + c = 0$
$ag + bf - c = 0$

AP EAMCET-23.08.2021

Conic Section

119897 The shortest distance from the line $3 x + 4 y = 25$ to the circle $x^{2} + y^{2} - 6 x + 8 y = 0$ is

1

\frac{9}{5}

2

\frac{7}{5}

3

\frac{8}{5}

4

\frac{13}{5}

Explanation:

B Given, equation of circle is
$x^{2} + y^{2} - 6 x + 8 y = 0$
$(x - 3)^{2} + (y + 4)^{2} = 25$
shortest distance $(d) =$ distance of line from centre - Radius
$= \frac{32}{5} - 5 = \frac{32 - 25}{5} d = \frac{7}{5}$

AP EAMCET-08.07.2022

Conic Section

119898 A circle touches the Y-axis at the point $(0, 4)$ and passes through the point $(2, 0)$ . Which of the following lines is not a tangent to this circle?

1

4 x - 3 y + 17 = 0

2

3 x + 4 y - 6 = 0

3

4 x + 3 y - 8 = 0

4

3 x - 4 y - 24 = 0

Explanation:

C Given point $(0, 4)$
Equation of family of circle
$(x - 0)^{2} + (y - 4)^{2} + λ x = 0$ passed through $(2, 0)$
$∴ (2 - 0)^{2} + (0 - 4)^{2} + 2 λ = 0$
$2^{2} + (- 4)^{2} + 2 λ = 0$
$4 + 16 + 2 λ = 0$
$λ = - 10$
Equation of circle
$x^{2} + y^{2} + 16 - 8 y + (- 10 x) = 0$
$x^{2} + y^{2} - 10 x - 8 y + 16 = 0$
centre, $(+ 5, 4)$
Radius $= \sqrt{25 + 16 - 16} = \sqrt{25} = 5$
$\Rightarrow$ checking option $A$
tangent $4 x + 3 y - 8 = 0$ distance from centre $= 5$
$| \frac{(4 \times 5) + (3 \times 4) - 8}{\sqrt{(4)^{2} + (3)^{2}}} | = | \frac{20 + 12 - 8}{5} | = | \frac{24}{5} | \neq 5$
$∴ 4 x + 3 y - 8 = 0$ not a tangent to circle.

JEE Main 09.01.2020

Conic Section

119895 The angle between the two tangents drawn from origin to the circle $x^{2} + y^{2} - 14 x + 2 y + 25$ $= 0$

1

0^{\circ}

2

45^{\circ}

3

90^{\circ}

4

60^{\circ}

Explanation:

C : original image
If $2 α$ is the angle between the two tangents from the origin to the circle.
$\tan α = \frac{{CT}_{1}}{{OT}_{1}} = \frac{\sqrt{49 + 1 - 25}}{\sqrt{25}} = 1$
$α = \tan^{- 1} (1) = 45^{\circ}$
$2 α = 2 \times 45^{\circ} = 90^{\circ} = \frac{π}{2}$ Angle between the tangents is $\frac{π}{2}$

AP EAMCET-23.08.2021

Conic Section

119896 The line $a x + b y + c = 0$ is normal to the circle $x^{2} + y^{2} + 2 g x + 2 f y + d = 0$ if

1

a g + b f + c = 0

2

\overset{―}{a g + b} f - c = 0

3

ag - bf + c = 0

4 ag

- bf - c = 0

Explanation:

B The centre of given circle is $(- g, - f)$ . If the
given line $ax + bx + c = 0$ is normal to the circle, then it
passes through the centre of circle
$∴ a (- g) + b (- f) + c = 0$
$ag + bf - c = 0$

AP EAMCET-23.08.2021

Conic Section

119897 The shortest distance from the line $3 x + 4 y = 25$ to the circle $x^{2} + y^{2} - 6 x + 8 y = 0$ is

1

\frac{9}{5}

2

\frac{7}{5}

3

\frac{8}{5}

4

\frac{13}{5}

Explanation:

B Given, equation of circle is
$x^{2} + y^{2} - 6 x + 8 y = 0$
$(x - 3)^{2} + (y + 4)^{2} = 25$
shortest distance $(d) =$ distance of line from centre - Radius
$= \frac{32}{5} - 5 = \frac{32 - 25}{5} d = \frac{7}{5}$

AP EAMCET-08.07.2022

Conic Section

119898 A circle touches the Y-axis at the point $(0, 4)$ and passes through the point $(2, 0)$ . Which of the following lines is not a tangent to this circle?

1

4 x - 3 y + 17 = 0

2

3 x + 4 y - 6 = 0

3

4 x + 3 y - 8 = 0

4

3 x - 4 y - 24 = 0

Explanation:

C Given point $(0, 4)$
Equation of family of circle
$(x - 0)^{2} + (y - 4)^{2} + λ x = 0$ passed through $(2, 0)$
$∴ (2 - 0)^{2} + (0 - 4)^{2} + 2 λ = 0$
$2^{2} + (- 4)^{2} + 2 λ = 0$
$4 + 16 + 2 λ = 0$
$λ = - 10$
Equation of circle
$x^{2} + y^{2} + 16 - 8 y + (- 10 x) = 0$
$x^{2} + y^{2} - 10 x - 8 y + 16 = 0$
centre, $(+ 5, 4)$
Radius $= \sqrt{25 + 16 - 16} = \sqrt{25} = 5$
$\Rightarrow$ checking option $A$
tangent $4 x + 3 y - 8 = 0$ distance from centre $= 5$
$| \frac{(4 \times 5) + (3 \times 4) - 8}{\sqrt{(4)^{2} + (3)^{2}}} | = | \frac{20 + 12 - 8}{5} | = | \frac{24}{5} | \neq 5$
$∴ 4 x + 3 y - 8 = 0$ not a tangent to circle.

JEE Main 09.01.2020

Conic Section

119895 The angle between the two tangents drawn from origin to the circle $x^{2} + y^{2} - 14 x + 2 y + 25$ $= 0$

1

0^{\circ}

2

45^{\circ}

3

90^{\circ}

4

60^{\circ}

Explanation:

C : original image
If $2 α$ is the angle between the two tangents from the origin to the circle.
$\tan α = \frac{{CT}_{1}}{{OT}_{1}} = \frac{\sqrt{49 + 1 - 25}}{\sqrt{25}} = 1$
$α = \tan^{- 1} (1) = 45^{\circ}$
$2 α = 2 \times 45^{\circ} = 90^{\circ} = \frac{π}{2}$ Angle between the tangents is $\frac{π}{2}$

AP EAMCET-23.08.2021

Conic Section

119896 The line $a x + b y + c = 0$ is normal to the circle $x^{2} + y^{2} + 2 g x + 2 f y + d = 0$ if

1

a g + b f + c = 0

2

\overset{―}{a g + b} f - c = 0

3

ag - bf + c = 0

4 ag

- bf - c = 0

Explanation:

B The centre of given circle is $(- g, - f)$ . If the
given line $ax + bx + c = 0$ is normal to the circle, then it
passes through the centre of circle
$∴ a (- g) + b (- f) + c = 0$
$ag + bf - c = 0$

AP EAMCET-23.08.2021

Conic Section

119897 The shortest distance from the line $3 x + 4 y = 25$ to the circle $x^{2} + y^{2} - 6 x + 8 y = 0$ is

1

\frac{9}{5}

2

\frac{7}{5}

3

\frac{8}{5}

4

\frac{13}{5}

Explanation:

B Given, equation of circle is
$x^{2} + y^{2} - 6 x + 8 y = 0$
$(x - 3)^{2} + (y + 4)^{2} = 25$
shortest distance $(d) =$ distance of line from centre - Radius
$= \frac{32}{5} - 5 = \frac{32 - 25}{5} d = \frac{7}{5}$

AP EAMCET-08.07.2022

Conic Section

119898 A circle touches the Y-axis at the point $(0, 4)$ and passes through the point $(2, 0)$ . Which of the following lines is not a tangent to this circle?

1

4 x - 3 y + 17 = 0

2

3 x + 4 y - 6 = 0

3

4 x + 3 y - 8 = 0

4

3 x - 4 y - 24 = 0

Explanation:

C Given point $(0, 4)$
Equation of family of circle
$(x - 0)^{2} + (y - 4)^{2} + λ x = 0$ passed through $(2, 0)$
$∴ (2 - 0)^{2} + (0 - 4)^{2} + 2 λ = 0$
$2^{2} + (- 4)^{2} + 2 λ = 0$
$4 + 16 + 2 λ = 0$
$λ = - 10$
Equation of circle
$x^{2} + y^{2} + 16 - 8 y + (- 10 x) = 0$
$x^{2} + y^{2} - 10 x - 8 y + 16 = 0$
centre, $(+ 5, 4)$
Radius $= \sqrt{25 + 16 - 16} = \sqrt{25} = 5$
$\Rightarrow$ checking option $A$
tangent $4 x + 3 y - 8 = 0$ distance from centre $= 5$
$| \frac{(4 \times 5) + (3 \times 4) - 8}{\sqrt{(4)^{2} + (3)^{2}}} | = | \frac{20 + 12 - 8}{5} | = | \frac{24}{5} | \neq 5$
$∴ 4 x + 3 y - 8 = 0$ not a tangent to circle.

JEE Main 09.01.2020