1.1 ANALYTICAL GEOMETRY II

EQUATION OF CIRCLE

Definition:

A circle is the locus of a point which moves in a plane in such a way that its distance from a fixed point remains constant. The fixed point is called the centre of the circle and the constant distance is called the radius of the circle.

Equation of the circle with centre (h, k) and radius ‘r’ units.

CP = r

√(( x – h )² + (y – k )² = r (Using distance formula)

(x – h )² + (y – k )² = r²

Note:

The equation of the circle with centre (0, 0 ) and radius ‘r’ units is x²+ y² = r²

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Example 1:

\[\color{Red}{What\ is\ the\ equation\ of\ the\ circle\ with\ centre\ at\ origin\ and\ radius\ 1\ unit?}\ \hspace{15cm}\]

\[Soln:\ \hspace{20cm}\]

\[We\ know\ that\ the\ equation\ of\ circle\ is\ (x\ -\ h)^2\ +\ (y\ -\ k)^2\ =\ r^2\]

\[Given\ h\ =\ 0,\ k = 0\ and\ r\ =\ 1\]

\[(x\ -\ 0)^2\ +\ (y\ -\ 0)^2\ =\ 1^2\]

\[x^2\ +\ y^2\ =\ 1\]

\[The\ equation\ of\ the\ circle\ is\ \boxed{x^2\ +\ y^2\ =\ 1}\]

Example 2: Find the equation of the circle with centre (- 5, 7 ) and radius 5 units.

Soln: We know that the equation of circle is (x – h )² + (y – k )² = r²

Here h = – 5, k = 7 (given) and r = 5

. . (x + 5 )² + (y + 7 )² = 5²

x² + 10x + 25 + y² + 14y + 49 = 25

x² + y² + 10x – 14y + 25 +49-25 = 0

x² + y² + 10x – 14y + 49 = 0

Therefore the equation of circle is x² + y²+ 10x – 14y + 49 = 0

General equation of the circle: x² + y² + 2gx + 2fy + c = 0

Centre = (-g , -f ) and radius r = √( g² + f² – c)

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Example 3: Find the centre and radius of the circle x² + y² + 2x + 2y – 7 = 0 .

Soln: Given x² + y² + 2x + 2y – 7 = 0 .

We know the equation of circle is x² + y² + 2gx + 2fy + c = 0

2g = 2 2f = 2 c = -7

g = 1 f = 1

centre = ( -g , -f ) r = √( g² + f² – c)

= ( – 1 , – 1 ) r = √( (1)² + (1)² + 7)

r = √9 r = 3

centre = (- 1 , – 1 ) & r = 3

Example 4:

\[\color {red}{Find\ the\ centre\ and\ radius\ of\ the\ circle\ x^2\ +\ y^2\ -\ 4\ x\ +\ 8\ y\ -\ 5\ =\ 0}\ \hspace{5cm}\]

\[ Soln:\ Given\ x^2\ +\ y^2\ -\ 4\ x\ +\ 8\ y\ -\ 5\ =\ 0\ \hspace{10cm}\]

\[We\ know\ that\ the\ equation\ of\ circle\ is\ x^2\ +\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{10cm}\]

\[2\ g\ =\ -\ 4\ \hspace{3cm}\ 2\ f\ =\ 8\ \hspace{3cm}\ c\ =\ -\ 5\]

\[g\ =\ -\ 2\ \hspace{3cm}\ f\ =\ 4\ \hspace{3cm}\]

\[centre\ =\ (-\ g,\ -\ f)\ \hspace{4cm}\ r\ =\ \sqrt{g^2\ +\ f^2\ -\ c}\]

\[centre\ =\ (2,\ -\ 4)\ \hspace{4cm}\ r\ =\ \sqrt{(-\ 2)^2\ +\ 4^2\ +\ 5}\]

\[\hspace{6cm}\ r\ =\ \sqrt{4\ +\ 16\ +\ 5}\]

\[\hspace{6cm}\ r\ =\ \sqrt{25}\]

\[\fbox{centre = (2, – 4) r = 5}\]

Example 5: Find the centre and radius of the circle x² + y² – 8y + 3 = 0 .

Soln: Given x² + y² – 8y + 3 = 0 .

We know the equation of circle is x² + y² + 2gx + 2fy + c = 0

2g = 0 2f = -8 c = 3

g = 0 f = -4

centre = ( -g , -f ) r = √( g² + f² – c)

= ( 0 , 4 ) r = √( (0)² + (- 4)² – 3)

r = √13

centre = (- 1 , – 1 ) & r = 3

Example 6:

\[\color {Red}{Show\ that\ the\ straight\ line\ 4\ x-\ y\ =\ 17\ passes\ through}\ \hspace{15cm}\]\[\color{red}{the\ centre\ of\ the\ circle\ x^2\ +\ y^2\ -\ 8\ x\ +\ 2\ y\ +\ 3\ =\ 0}\ \hspace{5cm}\]

\[ Soln:\ Given\ x^2\ +\ y^2\ -\ 8\ x\ +\ 2\ y\ +\ 3\ =\ 0\ \hspace{10cm}\]

\[We\ know\ that\ the\ equation\ of\ circle\ is\ x^2\ +\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ \hspace{10cm}\]

\[2\ g\ =\ -\ 8\ \hspace{3cm}\ 2\ f\ =\ 2\ \hspace{3cm}\ c\ =\ 3\]

\[g\ =\ -\ 4\ \hspace{3cm}\ f\ =\ 1\ \hspace{3cm}\]

\[centre\ =\ (-\ g,\ -\ f)\]

\[centre\ =\ (4,\ -\ 1)\]

\[4x\ -\ y=\ 17\ ——–(1)\ \hspace{10cm}\]

\[put\ x\ =\ 4,\ y\ =\ -\ 1\ in\ equation\ (1)\]

\[4(4)\ – (-1)\ =\ 17\]

\[17\ =\ 17\]

\[(4, -1)\ satisfies\ (1)\]

\[Hence\ the\ straight\ line\ 4\ x-\ y\ =\ 17\ passes\ through\ \hspace{15cm}\]\[the\ centre\ of\ the\ circle\ x^2\ +\ y^2\ -\ 8\ x\ +\ 2\ y\ +\ 3\ =\ 0\]

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Example 7: Find the equation of the circle passing through the point A (2, – 3) and having its centre at

C ( – 5 , 1).

Soln: r = √(( – 5 – 2 )² + (1 + 3 )²)

= √( (- 7)² + (4)² )

= √ ( 49 + 16)

r = √65

We know that the equation of circle is (x – h )² + (y – k )² = r²

Here h = -5, k = 1 (given) and r = √65

(x + 5 )² + (y – 1 )² = (√65) ²

x²+ 10x + 25 + y² – 2y + 1 = 65

x² + y² + 10x – 2y + 26 – 65 = 0

x² + y² + 10x – 2y – 39 = 0

Therefore the equation of circle is x² + y² + 10x – 2y – 39 = 0

Equation of circle on the line joining the points ( x₁, y₁) and ( x₂, y₂) as diameter :

\[(x\ -\ x_1)(x\ -\ x_2)\ +\ (y\ -\ y_1)(y\ -\ y_2)\ =\ 0\]

Example 1:

\[\color {red}{Find\ the\ equation\ of\ the\ circle\ on\ the\ line\ joining\ the\ points\ (2,3),\ (-\ 4,\ 5)\ as\ diameter.}\]\[\color{red}{Aslo\ find\ the\ centre\ and\ radius\ of\ the\ circle}\ \hspace{5cm}\]

\[Soln:\ \hspace{18cm}\]

\[(x\ -\ x_1)(x\ -\ x_2)\ +\ (y\ -\ y_1)(y\ -\ y_2)\ =\ 0\]

\[Given\ x_1\ =\ 2,\ y_1\ =\ 3,\ x_2\ =\ -\ 4,\ y_2\ =\ 5\]

\[(x\ -\ 2)(x\ +\ 4)\ +\ (y\ -\ 3)(y\ -\ 5)\ =\ 0\]

\[x^2\ +\ 4\ x\ – 2\ x\ – \ 8\ +\ y^2\ -\ 5y\ -\ 3y\ +\ 15\ =\ 0\]

\[x^2\ +\ 2\ x\ +\ y^2\ -\ 8y\ +\ 7\ =\ 0\]

\[x^2\ +\ y^2\ +\ 2\ x\ -\ 8y\ +\ 7\ =\ 0\]

\[\therefore\ the\ equation\ of\ the\ circle\ is\ \hspace{7cm}\]

\[\boxed{x^2\ +\ y^2\ +\ 2\ x\ -\ 8\ y\ +\ 7\ =\ 0}\ \hspace{5cm}\]

\[comparing\ with\ x^2\ +\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\]

\[2\ g\ =\ 2\ \hspace{3cm}\ 2\ f\ =\ -\ 8\ \hspace{3cm}\ c\ =\ 7\]

\[g\ =\ 1\ \hspace{3cm}\ f\ =\ -\ 4\ \hspace{3cm}\]

\[centre\ =\ (-\ g,\ -\ f)\ \hspace{4cm}\ r\ =\ \sqrt{g^2\ +\ f^2\ -\ c}\]

\[centre\ =\ (-\ 1,\ 4)\ \hspace{4cm}\ r\ =\ \sqrt{1^2\ +\ (-4)^2\ +\ 1}\]

\[\hspace{6cm}\ r\ =\ \sqrt{1\ +\ 16\ +\ 1}\]

\[\hspace{6cm}\ r\ =\ \sqrt{18}\]

\[\hspace{6cm}\ r\ =\ 3\sqrt{2}\]

\[\boxed{centre = (- 1, 4)\ \hspace{4cm}\ r = 3\sqrt{2}}\]

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FAMILY OF CIRCLES

Concentric Circles:

Two or more circles having the same centre but differ in radii are called concentric circles.

Note: Equation of concentric circles differ only by the constant term.

√ Concentric Circles (Definition and Example) | Σ Tricks

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Example 1 :Show that the circles x² + y² – 4x + 2y +5 = 0 and x² + y² – 4x + 2y – 8 = 0

are concentric circles.

Soln: From the two given equations of the circles, we observe that the constant term alone differs

∴The given circles are concentric circles.

Example 2 : Find the equation of the circle passing through the point (5 , 4 ) and concentric to the circle

x² + y² – 8x + 12y + 15 = 0 .

Soln: Equation of concentric circle be x² + y² – 8x + 12y + k = 0 ………….. ( 1 )

Put x = 5, y = 4 in ( 1 )

(5)² + (4)² – 8 ( 5 ) + 12 ( 4 ) + k = 0.

25 + 16 – 40 + 48 + k = 0.

89 – 40 + k = 0.

49 + k = 0

K = – 49

The required equation of the circle is x² + y² – 8x + 12y – 49 = 0

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Orthogonal Circles:

Two circles x² + y² + 2g₁x +2f₁y+c₁ = 0 and x² + y² + 2g₂x + 2f₂y + c₂ = 0 are said to be Orthogonal Circles if

2g₁g₂ + 2 f₁f₂ = c₁ + c₂

Part – B

Example 1:Prove that the circles x² + y² – 4x + 6y + 4 = 0 and

x² + y² + 2x + 4y + 4 = 0 cut orthogonally.

Soln: Given x² + y² – 4x + 6y + 4 = 0 ––––– (1) and

x² + y² + 2x + 4y + 4 = 0 ––––– (2)

From ( 1 )

2 g₁ = – 4 2 f₁ = 6 c₁ = 4

g₁= – 2 f₁= 3

From ( 2 )

2 g₂ =2 2 f₂ = 4 c₂ = 4

g₂= 1 f₂= 2

The condition for orthogonally is

2g₁g₂ + 2 f₁f₂ = c₁ + c₂

2(-2)(1) + 2(3)(2) = 4+ 4

-4 + 12 = 8

8 = 8

∴ The circles cut orthogonally.

Example 2:

\[\color{red}{Find\ the\ equation\ of\ the\ circle\ passing\ through\ the\ origin\ and\ cuts\ orthogonally}\ \hspace{3cm}\]\[\color{red}{each\ of\ the\ circles\ x^2\ +\ y^2\ -\ 6\ x\ +\ 8\ =\ 0\ and\ x^2\ +\ y^2\ -\ 2\ x\ -\ 2\ y\ -\ 7\ =\ 0}\]

\[Soln:\ \hspace{20cm}\]

\[Let\ the\ equation\ of\ circle\ be\ x^2\ +\ y^2\ +\ 2\ g\ x\ +\ 2\ f\ y\ +\ c\ =\ 0\ ——-(1)\]

\[(1)\ passes through (0,\ 0)\]

\[(0)^2\ +\ (0)^2\ +\ 2\ g(0)\ +\ 2f(0)\ +\ c\ =\ 0\]

\[c\ =\ 0\]

\[Given\ \hspace{10cm}\]

\[x^2\ +\ y^2\ -\ 6\ x\ +\ 8\ =\ 0 ——— (2)\]

\[x^2\ +\ y^2\ -\ 2\ x\ -\ 2\ y\ -\ 7\ =\ 0——— (3)\]

\[From\ (2)\ \hspace 10cm\]

\[2g_1 = -\ 6\ \hspace 2cm\ 2f_1 = 0\ \hspace 2cm\ c_1 = 8\]

\[g_1 =\ -\ 3\ \hspace 2cm\ f_1 = 0\ \hspace 2cm\ c_1 =\ 8\]

\[From\ (3)\ \hspace 10cm\]

\[2g_2 = -2\ \hspace 2cm\ 2f_2 = -2\ \hspace 2cm\ c_2 = – 7\]

\[g_2 = -1\ \hspace 2cm\ f_2 = -1\ \hspace 2cm\ c_2 = – 7\]

\[Given\ Equation\ (1)\ is\ orthogonal\ with\ x^2\ +\ y^2\ -\ 6\ x\ +\ 8\ =\ 0\]

\[2\ g\ g_1\ +\ 2\ f\ f_1\ =\ c\ +\ c_1\]

\[2\ g(-\ 3)\ +\ 2\ f (0)\ =\ c\ +\ 8\]

\[-\ 6g\ +\ 0\ =\ 0\ +\ 8\]

\[-\ 6g\ =\ 8\]

\[g\ =\ -\ \frac{8}{6}\]

\[\boxed{g\ =\ -\ \frac{4}{3}}\]

\[Given\ Equation\ (1)\ is\ orthogonal\ with\ x^2\ +\ y^2\ -\ 2\ x\ -\ 2\ y\ -\ 7\ =\ 0\]

\[2\ g\ g_2\ +\ 2\ f\ f_2\ =\ c\ +\ c_2\]

\[2\ g(-\ 1)\ +\ 2\ f (- 1)\ =\ c\ -\ 7\]

\[-\ 2g\ -\ 2\ f\ =\ 0\ -\ 7\]

\[-\ 2(-\ \frac{4}{3})\ -\ 2\ f\ =\ -\ 7\]

\[\frac{8}{3}\ -\ 2\ f\ =\ -\ 7\]

\[2\ f\ =\ \frac{8}{3}\ +\ 7\]

\[2\ f\ =\ \frac{8\ +\ 21}{3}\]

\[2f\ =\ \frac{29}{3}\]

\[\boxed{f\ =\ \frac{29}{6}}\]

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Contact of Circles:

Family of Circles - Study Material for IIT JEE | askIITians

Case ( i ) :

Two circles touch externally if the distance between their centres is equal to sum of their radii.

i.e C₁C₂= r₁ + r₂

Co-ordinates of point of contact are

P = ( (r₁x₂ + r₂ x₁) / (r₁ + r₂ ) , (r₁y₂ + r₂ y₁) / (r₁ + r₂ ) )

where C₁ = (x₁, y₁) and C₂ = (x₂, y₂)

Case ( ii ) :

Learn Common Tangents To Two Circles meaning, concepts, formulas through Study Material, Notes – Embibe.com

Two circles touch internally if the distance between their centres is equal to difference of their radii.

i.e C₁C₂= r₁ – r₂or r₂ – r₁

Co-ordinates of point of contact are

P = ( (r₁x₂ – r₂ x₁) / (r₁ – r₂ ) , (r₁y₂ – r₂ y₁) / (r₁ – r₂ ) )

where C₁ = (x₁, y₁) and C₂ = (x₂, y₂)

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Part – C

Example 1. Prove that the circles x² + y² + 2x – 4y – 3 = 0 and x² + y²– 8x + 6y +7 = 0 touch each other.

Soln: Given x² + y² + 2x – 4y – 3 = 0 ––––– (1) and

x² + y²– 8x + 6y +7 = 0 ––––– (2)

From ( 1 )

2g₁ = 2 2f₁ = – 4 c₁ = -3

g₁= 1 f₁= – 2

centre is C₁ = ( – g₁ , – f₁ ) r₁= √( g₁² + f₁² – c)

= ( -1 , 2 ) r₁= √( (1)² + (-2)² + 3)

r₁= √( 1 + 4 +3 )

r₁= √8 = 2√2

C₁= (-1 , 2) & r₁= 2√2

From ( 2 )

2g₂ = -8 2f₂ = 6 c₂ = 7

g₂ = – 4 f₂= 3

centre is C₂ = ( – g₂ , – f₂ ) r₂= √( g₂² + f₂² – c)

= ( 4 , -3 ) r₂= √( (-4)² + (3)² – 7)

r₂= √( 16 + 9 – 7 )

r₂= √18 = √(9 × 2) = 3√2

C₂= ( 4 , -3 ) & r₂= 3√2

C₁C₂ = √(( – 1 – 4 )² + (2 + 3 )²)

= √( (- 5)² + (5)² )

= √( 25 + 25)

= √50 = √(25 × 2) = 5√2

C₁C₂ = 5√2

r₁ + r₂ = 2√2 + 3√2

= 5√2 = C₁C₂

∴The circles touch each other externally.

Example 2. Prove that the circles x² + y² – 4x + 6y – 112= 0 and x² + y² – 10x – 6y + 14 = 0 touch each other.

Soln: Given x² + y² – 4x + 6y – 112 = 0 ––––– (1) and

x² + y² – 10x – 6y + 14 = 0 ––––– (2)

From ( 1 )

2g₁ = -4 2f₁ = 6 c₁ = -112

g₁= -2 f₁= 3

centre is C₁ = ( – g₁ , – f₁ ) r₁= √( g₁² + f₁² – c)

= ( 2 , -3 ) r₁= √( (-2)² + (3)² + 112)

r₁= √( 4 + 9 + 112 )

r₁= √125 = √(125 × 5) = 5√5

C₁= ( 2 , -3 ) & r₁= 5√5

From ( 2 )

2 g₂ = – 10 2 f₂ = -6 c₂ = 14

g₂ = -5 f₂ = – 3

centre is C₂ = ( – g₂ , – f₂ ) r₂= √( g₂² + f₂² – c)

= ( 5 , 3 ) r₂= √( (-5)² + (-3)² – 14)

r₂= √( 25 + 9 – 14 )

r₂ = √20 = √(4 × 5)= 2√5

C₂= ( 5 , 3 ) & r₂= 2√5

C₁ C₂ = √(( 5 – 2 )² + (3 + 3 )²)

= √( (3)² + (6)² )

= √( 9 + 36)

C₁ C₂ = √45 = √(9× 5) = 3√5

r₁ – r₂ = 5√5 – 2√5

= 3√5 = C₁ C₂

∴The circles touch each other internally.

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