QBManager

Conic Section

119752 For the circle $x^{2} + y^{2} - 2 x - 4 y + 3 = 0$ , the point

1

(0, 1)

lies on the circle
Prove,

x = 0, y = 1

0 + 1 - 0 - 4 + 3 = 0

4 - 4 = 0 (lies on the circle)

2

(3, 1)

lies outside the circle
Prove,

x^{2} + y^{2} - 2 x - 4 y + 3 = 0

9 + 1 - 6 - 4 + 3 = 0

3 > 0 (lies outside the circle)

3

(1, 3)

lies inside the circle
Prove,

x^{2} + y^{2} - 2 x - 4 y + 3 = 0

1 + 9 - 2 - 12 + 3 = 0

- 1 < 0 (lies inside the circle)

4

(1, 1)

lies outside the circle
Prove,

x^{2} + y^{2} - 2 x - 4 y + 3 = 0

1 + 1 - 2 - 4 + 3 = 0

- 1 < 0 (lies inside the circle)

- 1 < 0

(lies inside the circle)
So, a, b, c are correct.

Explanation:

A $x^{2} + y^{2} - 2 x - 4 y + 3 = 0$
(a.) $(0, 1)$ lies on the circle
Prove,
$x = 0, y = 1$
$0 + 1 - 0 - 4 + 3 = 0$
$4 - 4 = 0 (lies on the circle)$
(b.) $(3, 1)$ lies outside the circle
Prove,
$x^{2} + y^{2} - 2 x - 4 y + 3 = 0$
$9 + 1 - 6 - 4 + 3 = 0$
$3 > 0 (lies outside the circle)$
(c.) $(1, 3)$ lies inside the circle
Prove,
$x^{2} + y^{2} - 2 x - 4 y + 3 = 0$
$1 + 9 - 2 - 12 + 3 = 0$
$- 1 < 0 (lies inside the circle)$
(d.) $(1, 1)$ lies outside the circle
Prove,
$x^{2} + y^{2} - 2 x - 4 y + 3 = 0$
$1 + 1 - 2 - 4 + 3 = 0$
$- 1 < 0 (lies inside the circle)$
$- 1 < 0$ (lies inside the circle)
So, a, b, c are correct.

UPSEE-2011

Conic Section

119753 The angle between the tangent drawn from origin to the circle $(x - 7)^{2} + (y + 1)^{2} = 25$ is

1

\frac{π}{3}

2

\frac{π}{6}

3

\frac{π}{2}

4

\frac{π}{8}

Explanation:

C Let tangent from origin be $y = mx$ Using the condition of tangency we get.
$\frac{7 m + 1}{\sqrt{m^{2} + 1}} = 5$
$(7 m + 1)^{2} = {(5 \sqrt{m^{2} + 1})}^{2}$
$49 m^{2} + 1 + 14 m = 25 (m^{2} + 1)$
$49 m^{2} - 25 m^{2} + 14 m - 25 + 1 = 0$
$24 m^{2} + 14 m - 24 = 0$
$12 m^{2} + 7 m - 12 = 0$
$12 m^{2} + 16 m - 9 m - 12 = 0$
$4 m (3 m + 4) - 3 (3 m + 4) = 0$
$(3 m + 4) (4 m - 3) = 0$
$m = \frac{- 4}{3}, m = \frac{3}{4} .$
Therefore the product of both the slopes is -1 , i.e.
$\frac{3}{4} \times \frac{- 4}{3} = - 1$ Hence the angle between the two tangent is $\frac{π}{2}$ .

UPSEE-2010

Conic Section

119754 $A B C D$ is a square of unit area. $A$ circle is tangent to two sides of $ABCD$ and passes through exactly one of its vertices. The radius of the circle is

1

2 - \sqrt{2}

2

\sqrt{2} - 1

3

\frac{1}{2}

4

\frac{1}{\sqrt{2}}

Explanation:

A $:$ Let the centre of the circle is $(r, r)$
$OE = CE = (1 - r)$
original image
Applying Pythagoras theorem
in are $Δ OED$
$(1 - r)^{2} + (1 - r)^{2} = r^{2}$
$2 (1 - r)^{2} = r^{2}$
$\sqrt{2} (1 - r) = r$
$r (\sqrt{2} + 1) = \sqrt{2}$
$r = \frac{\sqrt{2}}{\sqrt{2} + 1} = 2 - \sqrt{2}$

AMU-2019

Conic Section

119755 If the line $x + 2 b y + 7 = 0$ is a diameter of the circle $x^{2} + y^{2} - 6 x + 2 y = 0$ , then $b$ is equal to

1 -5

2 -3

3 2

4 5

Explanation:

D We have equation of circle is
$x^{2} + y^{2} - 6 x + 2 y = 0$
Ans: (7)
Exp: (7) : original image
Given, that
$k = r$
$h = 1$
$OP = r, PR = 1$
$OR = | \frac{r + 1}{\sqrt{2}} |$
$r^{2} = 1 + \frac{(r + 1)^{2}}{2}$
$2 r^{2} = 2 + r^{2} + 1 + 2 r$
$r^{2} - 2 r - 3 = 0$
$(r - 3) (r + 1) = 0$
$r = 3, - 1$
$h + k + r = 1 + 3 + 3 \Rightarrow 7$

JEE Main-24.06.2022

Conic Section

119752 For the circle $x^{2} + y^{2} - 2 x - 4 y + 3 = 0$ , the point

1

(0, 1)

lies on the circle
Prove,

x = 0, y = 1

0 + 1 - 0 - 4 + 3 = 0

4 - 4 = 0 (lies on the circle)

2

(3, 1)

lies outside the circle
Prove,

x^{2} + y^{2} - 2 x - 4 y + 3 = 0

9 + 1 - 6 - 4 + 3 = 0

3 > 0 (lies outside the circle)

3

(1, 3)

lies inside the circle
Prove,

x^{2} + y^{2} - 2 x - 4 y + 3 = 0

1 + 9 - 2 - 12 + 3 = 0

- 1 < 0 (lies inside the circle)

4

(1, 1)

lies outside the circle
Prove,

x^{2} + y^{2} - 2 x - 4 y + 3 = 0

1 + 1 - 2 - 4 + 3 = 0

- 1 < 0 (lies inside the circle)

- 1 < 0

(lies inside the circle)
So, a, b, c are correct.

Explanation:

A $x^{2} + y^{2} - 2 x - 4 y + 3 = 0$
(a.) $(0, 1)$ lies on the circle
Prove,
$x = 0, y = 1$
$0 + 1 - 0 - 4 + 3 = 0$
$4 - 4 = 0 (lies on the circle)$
(b.) $(3, 1)$ lies outside the circle
Prove,
$x^{2} + y^{2} - 2 x - 4 y + 3 = 0$
$9 + 1 - 6 - 4 + 3 = 0$
$3 > 0 (lies outside the circle)$
(c.) $(1, 3)$ lies inside the circle
Prove,
$x^{2} + y^{2} - 2 x - 4 y + 3 = 0$
$1 + 9 - 2 - 12 + 3 = 0$
$- 1 < 0 (lies inside the circle)$
(d.) $(1, 1)$ lies outside the circle
Prove,
$x^{2} + y^{2} - 2 x - 4 y + 3 = 0$
$1 + 1 - 2 - 4 + 3 = 0$
$- 1 < 0 (lies inside the circle)$
$- 1 < 0$ (lies inside the circle)
So, a, b, c are correct.

UPSEE-2011

Conic Section

119753 The angle between the tangent drawn from origin to the circle $(x - 7)^{2} + (y + 1)^{2} = 25$ is

1

\frac{π}{3}

2

\frac{π}{6}

3

\frac{π}{2}

4

\frac{π}{8}

Explanation:

C Let tangent from origin be $y = mx$ Using the condition of tangency we get.
$\frac{7 m + 1}{\sqrt{m^{2} + 1}} = 5$
$(7 m + 1)^{2} = {(5 \sqrt{m^{2} + 1})}^{2}$
$49 m^{2} + 1 + 14 m = 25 (m^{2} + 1)$
$49 m^{2} - 25 m^{2} + 14 m - 25 + 1 = 0$
$24 m^{2} + 14 m - 24 = 0$
$12 m^{2} + 7 m - 12 = 0$
$12 m^{2} + 16 m - 9 m - 12 = 0$
$4 m (3 m + 4) - 3 (3 m + 4) = 0$
$(3 m + 4) (4 m - 3) = 0$
$m = \frac{- 4}{3}, m = \frac{3}{4} .$
Therefore the product of both the slopes is -1 , i.e.
$\frac{3}{4} \times \frac{- 4}{3} = - 1$ Hence the angle between the two tangent is $\frac{π}{2}$ .

UPSEE-2010

Conic Section

119754 $A B C D$ is a square of unit area. $A$ circle is tangent to two sides of $ABCD$ and passes through exactly one of its vertices. The radius of the circle is

1

2 - \sqrt{2}

2

\sqrt{2} - 1

3

\frac{1}{2}

4

\frac{1}{\sqrt{2}}

Explanation:

A $:$ Let the centre of the circle is $(r, r)$
$OE = CE = (1 - r)$
original image
Applying Pythagoras theorem
in are $Δ OED$
$(1 - r)^{2} + (1 - r)^{2} = r^{2}$
$2 (1 - r)^{2} = r^{2}$
$\sqrt{2} (1 - r) = r$
$r (\sqrt{2} + 1) = \sqrt{2}$
$r = \frac{\sqrt{2}}{\sqrt{2} + 1} = 2 - \sqrt{2}$

AMU-2019

Conic Section

119755 If the line $x + 2 b y + 7 = 0$ is a diameter of the circle $x^{2} + y^{2} - 6 x + 2 y = 0$ , then $b$ is equal to

1 -5

2 -3

3 2

4 5

Explanation:

D We have equation of circle is
$x^{2} + y^{2} - 6 x + 2 y = 0$
Ans: (7)
Exp: (7) : original image
Given, that
$k = r$
$h = 1$
$OP = r, PR = 1$
$OR = | \frac{r + 1}{\sqrt{2}} |$
$r^{2} = 1 + \frac{(r + 1)^{2}}{2}$
$2 r^{2} = 2 + r^{2} + 1 + 2 r$
$r^{2} - 2 r - 3 = 0$
$(r - 3) (r + 1) = 0$
$r = 3, - 1$
$h + k + r = 1 + 3 + 3 \Rightarrow 7$

JEE Main-24.06.2022

Conic Section

119752 For the circle $x^{2} + y^{2} - 2 x - 4 y + 3 = 0$ , the point

1

(0, 1)

lies on the circle
Prove,

x = 0, y = 1

0 + 1 - 0 - 4 + 3 = 0

4 - 4 = 0 (lies on the circle)

2

(3, 1)

lies outside the circle
Prove,

x^{2} + y^{2} - 2 x - 4 y + 3 = 0

9 + 1 - 6 - 4 + 3 = 0

3 > 0 (lies outside the circle)

3

(1, 3)

lies inside the circle
Prove,

x^{2} + y^{2} - 2 x - 4 y + 3 = 0

1 + 9 - 2 - 12 + 3 = 0

- 1 < 0 (lies inside the circle)

4

(1, 1)

lies outside the circle
Prove,

x^{2} + y^{2} - 2 x - 4 y + 3 = 0

1 + 1 - 2 - 4 + 3 = 0

- 1 < 0 (lies inside the circle)

- 1 < 0

(lies inside the circle)
So, a, b, c are correct.

Explanation:

A $x^{2} + y^{2} - 2 x - 4 y + 3 = 0$
(a.) $(0, 1)$ lies on the circle
Prove,
$x = 0, y = 1$
$0 + 1 - 0 - 4 + 3 = 0$
$4 - 4 = 0 (lies on the circle)$
(b.) $(3, 1)$ lies outside the circle
Prove,
$x^{2} + y^{2} - 2 x - 4 y + 3 = 0$
$9 + 1 - 6 - 4 + 3 = 0$
$3 > 0 (lies outside the circle)$
(c.) $(1, 3)$ lies inside the circle
Prove,
$x^{2} + y^{2} - 2 x - 4 y + 3 = 0$
$1 + 9 - 2 - 12 + 3 = 0$
$- 1 < 0 (lies inside the circle)$
(d.) $(1, 1)$ lies outside the circle
Prove,
$x^{2} + y^{2} - 2 x - 4 y + 3 = 0$
$1 + 1 - 2 - 4 + 3 = 0$
$- 1 < 0 (lies inside the circle)$
$- 1 < 0$ (lies inside the circle)
So, a, b, c are correct.

UPSEE-2011

Conic Section

119753 The angle between the tangent drawn from origin to the circle $(x - 7)^{2} + (y + 1)^{2} = 25$ is

1

\frac{π}{3}

2

\frac{π}{6}

3

\frac{π}{2}

4

\frac{π}{8}

Explanation:

C Let tangent from origin be $y = mx$ Using the condition of tangency we get.
$\frac{7 m + 1}{\sqrt{m^{2} + 1}} = 5$
$(7 m + 1)^{2} = {(5 \sqrt{m^{2} + 1})}^{2}$
$49 m^{2} + 1 + 14 m = 25 (m^{2} + 1)$
$49 m^{2} - 25 m^{2} + 14 m - 25 + 1 = 0$
$24 m^{2} + 14 m - 24 = 0$
$12 m^{2} + 7 m - 12 = 0$
$12 m^{2} + 16 m - 9 m - 12 = 0$
$4 m (3 m + 4) - 3 (3 m + 4) = 0$
$(3 m + 4) (4 m - 3) = 0$
$m = \frac{- 4}{3}, m = \frac{3}{4} .$
Therefore the product of both the slopes is -1 , i.e.
$\frac{3}{4} \times \frac{- 4}{3} = - 1$ Hence the angle between the two tangent is $\frac{π}{2}$ .

UPSEE-2010

Conic Section

119754 $A B C D$ is a square of unit area. $A$ circle is tangent to two sides of $ABCD$ and passes through exactly one of its vertices. The radius of the circle is

1

2 - \sqrt{2}

2

\sqrt{2} - 1

3

\frac{1}{2}

4

\frac{1}{\sqrt{2}}

Explanation:

A $:$ Let the centre of the circle is $(r, r)$
$OE = CE = (1 - r)$
original image
Applying Pythagoras theorem
in are $Δ OED$
$(1 - r)^{2} + (1 - r)^{2} = r^{2}$
$2 (1 - r)^{2} = r^{2}$
$\sqrt{2} (1 - r) = r$
$r (\sqrt{2} + 1) = \sqrt{2}$
$r = \frac{\sqrt{2}}{\sqrt{2} + 1} = 2 - \sqrt{2}$

AMU-2019

Conic Section

119755 If the line $x + 2 b y + 7 = 0$ is a diameter of the circle $x^{2} + y^{2} - 6 x + 2 y = 0$ , then $b$ is equal to

1 -5

2 -3

3 2

4 5

Explanation:

D We have equation of circle is
$x^{2} + y^{2} - 6 x + 2 y = 0$
Ans: (7)
Exp: (7) : original image
Given, that
$k = r$
$h = 1$
$OP = r, PR = 1$
$OR = | \frac{r + 1}{\sqrt{2}} |$
$r^{2} = 1 + \frac{(r + 1)^{2}}{2}$
$2 r^{2} = 2 + r^{2} + 1 + 2 r$
$r^{2} - 2 r - 3 = 0$
$(r - 3) (r + 1) = 0$
$r = 3, - 1$
$h + k + r = 1 + 3 + 3 \Rightarrow 7$

JEE Main-24.06.2022

Conic Section

119752 For the circle $x^{2} + y^{2} - 2 x - 4 y + 3 = 0$ , the point

1

(0, 1)

lies on the circle
Prove,

x = 0, y = 1

0 + 1 - 0 - 4 + 3 = 0

4 - 4 = 0 (lies on the circle)

2

(3, 1)

lies outside the circle
Prove,

x^{2} + y^{2} - 2 x - 4 y + 3 = 0

9 + 1 - 6 - 4 + 3 = 0

3 > 0 (lies outside the circle)

3

(1, 3)

lies inside the circle
Prove,

x^{2} + y^{2} - 2 x - 4 y + 3 = 0

1 + 9 - 2 - 12 + 3 = 0

- 1 < 0 (lies inside the circle)

4

(1, 1)

lies outside the circle
Prove,

x^{2} + y^{2} - 2 x - 4 y + 3 = 0

1 + 1 - 2 - 4 + 3 = 0

- 1 < 0 (lies inside the circle)

- 1 < 0

(lies inside the circle)
So, a, b, c are correct.

Explanation:

A $x^{2} + y^{2} - 2 x - 4 y + 3 = 0$
(a.) $(0, 1)$ lies on the circle
Prove,
$x = 0, y = 1$
$0 + 1 - 0 - 4 + 3 = 0$
$4 - 4 = 0 (lies on the circle)$
(b.) $(3, 1)$ lies outside the circle
Prove,
$x^{2} + y^{2} - 2 x - 4 y + 3 = 0$
$9 + 1 - 6 - 4 + 3 = 0$
$3 > 0 (lies outside the circle)$
(c.) $(1, 3)$ lies inside the circle
Prove,
$x^{2} + y^{2} - 2 x - 4 y + 3 = 0$
$1 + 9 - 2 - 12 + 3 = 0$
$- 1 < 0 (lies inside the circle)$
(d.) $(1, 1)$ lies outside the circle
Prove,
$x^{2} + y^{2} - 2 x - 4 y + 3 = 0$
$1 + 1 - 2 - 4 + 3 = 0$
$- 1 < 0 (lies inside the circle)$
$- 1 < 0$ (lies inside the circle)
So, a, b, c are correct.

UPSEE-2011

Conic Section

119753 The angle between the tangent drawn from origin to the circle $(x - 7)^{2} + (y + 1)^{2} = 25$ is

1

\frac{π}{3}

2

\frac{π}{6}

3

\frac{π}{2}

4

\frac{π}{8}

Explanation:

C Let tangent from origin be $y = mx$ Using the condition of tangency we get.
$\frac{7 m + 1}{\sqrt{m^{2} + 1}} = 5$
$(7 m + 1)^{2} = {(5 \sqrt{m^{2} + 1})}^{2}$
$49 m^{2} + 1 + 14 m = 25 (m^{2} + 1)$
$49 m^{2} - 25 m^{2} + 14 m - 25 + 1 = 0$
$24 m^{2} + 14 m - 24 = 0$
$12 m^{2} + 7 m - 12 = 0$
$12 m^{2} + 16 m - 9 m - 12 = 0$
$4 m (3 m + 4) - 3 (3 m + 4) = 0$
$(3 m + 4) (4 m - 3) = 0$
$m = \frac{- 4}{3}, m = \frac{3}{4} .$
Therefore the product of both the slopes is -1 , i.e.
$\frac{3}{4} \times \frac{- 4}{3} = - 1$ Hence the angle between the two tangent is $\frac{π}{2}$ .

UPSEE-2010

Conic Section

119754 $A B C D$ is a square of unit area. $A$ circle is tangent to two sides of $ABCD$ and passes through exactly one of its vertices. The radius of the circle is

1

2 - \sqrt{2}

2

\sqrt{2} - 1

3

\frac{1}{2}

4

\frac{1}{\sqrt{2}}

Explanation:

A $:$ Let the centre of the circle is $(r, r)$
$OE = CE = (1 - r)$
original image
Applying Pythagoras theorem
in are $Δ OED$
$(1 - r)^{2} + (1 - r)^{2} = r^{2}$
$2 (1 - r)^{2} = r^{2}$
$\sqrt{2} (1 - r) = r$
$r (\sqrt{2} + 1) = \sqrt{2}$
$r = \frac{\sqrt{2}}{\sqrt{2} + 1} = 2 - \sqrt{2}$

AMU-2019

Conic Section

119755 If the line $x + 2 b y + 7 = 0$ is a diameter of the circle $x^{2} + y^{2} - 6 x + 2 y = 0$ , then $b$ is equal to

1 -5

2 -3

3 2

4 5

Explanation:

D We have equation of circle is
$x^{2} + y^{2} - 6 x + 2 y = 0$
Ans: (7)
Exp: (7) : original image
Given, that
$k = r$
$h = 1$
$OP = r, PR = 1$
$OR = | \frac{r + 1}{\sqrt{2}} |$
$r^{2} = 1 + \frac{(r + 1)^{2}}{2}$
$2 r^{2} = 2 + r^{2} + 1 + 2 r$
$r^{2} - 2 r - 3 = 0$
$(r - 3) (r + 1) = 0$
$r = 3, - 1$
$h + k + r = 1 + 3 + 3 \Rightarrow 7$

JEE Main-24.06.2022

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