SATHEE: CIRCLE-2 (Position of a Point)

CIRCLE-2 (Position of a Point)

1. Position of a point with respect to a circle .

let the circle is $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$

Point $P (x_{1}, y_{1})$ lies outside, on or inside the circle accordingly $CP >$ , $=$ , $<$ radius

(or) $S_{1} = x_{1}^{2} + y_{1}^{2} + 2 {gx}_{1} + 2 {fy}_{1} + c >, =, < 0$

2. Maximum and minimum distance of a point from the circle

Let the circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ and point $P (x_{1}, y_{1})$

The maximum and minimum distance form $P (x_{1}, y_{1})$ to the circle are

$PB = CB + CP$

$= r + CP$

$PA = | CP - CA | = | PC - r |$

$PB$ is maximum distance and $PA$ is minimum distance.

Examples

1. If the equation $p x^{2} + (3 - q) x y + 2 y^{2} - 6 q x + 30 y + 6 q = 0$ represents a circle, then the values of $p$ and q are

(a) 3, 1

(b) 2, 2

(d) 3, 4

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Solution :

${ax}^{2} + {by}^{2} + 2 hxy + 2 gx + 2 fy + c 0$

represents a circle if $h = 0$ and $a = b$

$∴ p = 2$ and $3 - q = 0 \Rightarrow q = 3$

Correct option is :(c)

2. The number if integral values of $λ$ for which $x^{2} + y^{2} + (1 - λ) x + λ y + 5 = 0$ is the equation of a circle whose radius cannot exceed 5 is

(a) 20

(b) 16

(d) 24

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Solution :

radius of the equation $\sqrt{g^{2} + f^{2} - c}$

$g = \frac{1 - λ}{2}, f = \frac{λ}{2} c = 5$

$\sqrt{{(\frac{1 - λ}{2})}^{2} + \frac{λ^{2}}{4} - 5} \leq 5$

$1 + λ^{2} - 2 λ + λ^{2} - 20 \leq 300$

$2 λ^{2} - 2 λ - 119 \leq 0$

$\begin{aligned} D = 4 + 8 \times 119 \\ = 4 + 952 \\ = 956 \\ = 4 (239) \\ λ = \frac{2 \pm 2 \sqrt{239}}{4} \\ = \frac{1 - \sqrt{239}}{2} \frac{1 + \sqrt{239}}{2} \\ ∴ \frac{1 - \sqrt{239}}{2} \leq λ \leq \frac{1 + \sqrt{239}}{2} \\ - 7.2 \leq λ \leq 8.2 (approx) \\ λ = - 7, - 6, - 5, - 4, \dots \dots . .8 \end{aligned}$

number of values of $λ$ is 16

3. The equation of the circle which passes through $(1, 0)$ and $(0, 1)$ and has its radius as small as possible is

(a) $x^{2} + y^{2} - 2 x - 2 y + 1 = 0$

(b) $x^{2} + y^{2} - x - y = 0$

(d) $x^{2} + y^{2} - 3 x - 3 y + 2 = 0$

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Solution :

The radius will be minimum, if the given points are the end points of a diameter. Then the equation of the circle is $(x - 1) (x - 0) + (y - 0) (y - 1) = 0 \Rightarrow x^{2} + y^{2} - x - y = 0$

4. The centre of the circle $x = - 1 + 2 \cos θ, y = 3 + 2 \sin θ$ is

(a) $(1, - 3)$

(b) $(- 1, 3)$

(d) None of these

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Solution :

Rewrite the given equation

$\begin{array}{ll} x + 1 = 2 \cos θ & y - 3 = 2 \sin θ \\ \frac{x + 1}{2} = \cos θ & \frac{y - 3}{2} = \sin θ \end{array}$

squaring and adding

$\begin{aligned} {(\frac{x + 1}{2})}^{2} + {(\frac{y - 3}{2})}^{2} = \cos^{2} θ + \sin^{2} θ \\ (x + 1)^{2} + (y - 3)^{2} = 4 \\ centre (- 1, 3) radius 2 \end{aligned}$

Circle

CIRCLE-2 (Position of a Point)

1. Position of a point with respect to a circle .

2. Maximum and minimum distance of a point from the circle

Examples

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Mentorship

NCERT Exercise Video Solution

Circle

CIRCLE-2 (Position of a Point)

1. Position of a point with respect to a circle .

2. Maximum and minimum distance of a point from the circle

Examples

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Mentorship

NCERT Exercise Video Solution

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