HOW TO SHOW TWO CIRCLES ARE ORTHOGONAL

Two circles are said to be orthogonal circles, if the tangent at their point of intersection are at right angles.

Example 1 :

Show that the circles

x² + y² - 8x + 6y - 23 = 0

and

x² + y² - 2x - 5y + 16 = 0

are orthogonal.

Solution :

From 1^st equation :

x² + y² - 8x + 6y - 23  =  0

x² + y² + 2g₁x + 2f₁y + c₁  =  0

2g₁ = -8, g₁ = -4 

2f₁ = 6, f₁ = 3 

and c₁ = -2

From 2^nd equation :

x² + y² - 2x - 5y + 16 = 0 

2g₂ = -2, g₂  =  -1

2f₂ = -5, f₂ = -5/2

and c₂ = 16

Condition to prove two circles are orthogonal :

2 g₁ g₂ + 2 f₁ f₂  =  c₁ + c₂

2(-4)(-1)+2(3)(-5/2)  =  -23+16

8-15  =  -7

- 7  =  -7

The two circles cut orthogonally and hence they are orthogonal circles.

Example 2 :

Show that the circles

x² + y² - 8x + 6y + 21  =  0

and

x² + y² - 2y - 15  =  0

are orthogonal.

Solution :

From 1^st equation :

x² + y² - 8x + 6y + 21 = 0

x² + y² + 2g₁x + 2f₁y + c₁  =  0

2g₁ = -8, g₁ = -4

2f₁ = 6, f₁ = 3 

and c₁ = 21

From 2^nd equation :

x² + y² - 2y - 15 = 0

x² + y² + 2g₂x + 2f₂y + c₂  =  0

2g₁ = 0, g₁ = 0

2f₁ = -2, f₁ = -1 

and c₁ = -15

2 (-4) (0) + 2 (-3) (-1) = 21 - 15

Condition to prove two circles are orthogonal :

2 g₁ g₂ + 2 f₁ f₂  =  c₁ + c₂

2(-4)(0)+2(-3)(-1)  =  21-15

0+6  =  6

6  =  6

The two circles cut orthogonally and hence they are orthogonal circles. 

Example 3 :

Find the equation of the circle which passes through the point (1, 2) and cuts orthogonally each of the circles

x² + y² = 9

and

x² + y² − 2x + 8y − 7 = 0

Solution :

Let the required equation of the circle be

x² + y² + 2gx + 2fy + c = 0 ------(1)

The point (1, 2) lies on the circle

1+4+2g+4f+c  =  0

2g+4f+c  =  −5 -----(2)

The circle (1) cuts the circle x² + y² = 9 orthogonally.

2g₁g₂ + 2f₁f₂ = c₁ + c₂

2g(0) + 2f(0) = c − 9

 c = 9 ---(3)

Again the circle (1) cuts 

x² + y² − 2x + 8y − 7 = 0

orthogonally.

2g(− 1) + 2f(4) = c − 7

-2g + 8f = 9 − 7 = 2

−g + 4f = 1 -----(4)

(2) becomes 2g + 4f = − 14

g + 2f = − 7 ------(5)

(4) + (5) => 6f = − 6 ⇒ f = − 1

(5) => g − 2 = − 7 => g = − 5

The required equation of the circle is

x² + y² − 10x − 2y + 9 = 0

Example 4 :

A circle S passes through the point (0, 1) and is orthogonal to the circles (x - 1)² + y² = 16 and x² + y² = 1. Then

a) radius of S is 8     b) radius of S is 7

c) center of S is (-7, 1)

d) center of S is (-8, 1)

Solution :

Let the required equation of the circle be

x² + y² + 2gx + 2fy + c = 0 ------(1)

The point (0, 1) lies on the circle

0 + 1 + 2g(0) + 2f(1) + c = 0

1 + 2f + c = 0

2f + c = -1 -----(2)

The circle (1) cuts the circle

(x - 1)² + y² = 16

orthogonally,

x² - 2x + 1 + y² = 16

x² + y² - 2x + 1 - 16 = 0

x² + y² - 2x - 15 = 0

2g₂ = -2, 2f₂ = 0

g₂ = -1, f₂ = 0, c₂ = -15 

2g₁g₂ + 2f₁f₂ = c₁ + c₂

2g(-1) + 2f(0) = c − 15

-2g  = c − 15

-2g - c = -15

2g + c = 15 -----(3)

The circle (1) cuts the circle

x² + y² = 1

orthogonally,

2g₂ = 0, 2f₂ = 0, c₂ = -1

g₂ = 0, f₂ = 0, c₂ = -1

2g₁g₂ + 2f₁f₂ = c₁ + c₂

2g(0) + 2f(0) = c − 1

c = 1

Applying the value of c in (3), we get

2g + 1 = 15

2g = 14

g = 7

Applying the values of g and c in (1), we get

x² + y² + 2(7)x + 2fy + 1 = 0

Since the required circle passes through the point (0, 1), we can apply.

0² + 1² + 2(7)(0) + 2f(1) + 1 = 0

1 + 2f + 1 = 0

2f = -2

f = -1

Applying into the equation, we get

x² + y² + 2(7)x + 2(-1)y + 1 = 0

x² + y² + 14x - 2y + 1 = 0

Center (-g, -f) ==> (-7, 1)

So, option c is correct.

Finding radius :

= √g² + f² - c

= √(-7)² + 1² - 1

= 7

Then, option b is also correct.

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