Circle (LAQs)

Maths-2B | 1. Circle – LAQs:
Welcome to LAQs in Chapter 1: Circle. This page contains the most Important FAQs for Long Answer Questions in this Chapter. Each answer is provided in simple and easy-to-understand steps. This will support your preparation and help you secure top marks in your exams.

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LAQ-1 : Find the equation of the circle passing through the points A(1,2), B(3,-4), C(5,-6)

Given $$A = (1,2) B = (3,-4) C = (5,-6)$$

We take $$S(x_1,y_1)$$ as the centre of the circle

$$\Rightarrow SA = SB = SC$$

Now $$SA = SB$$

$$\Rightarrow SA^2 = SB^2$$

$$\Rightarrow (x_1-1)^2+(y_1-2)^2 = (x_1-3)^2+(y_1+4)^2$$

$$\Rightarrow (x_1^2 – 2x_1 + 1) + (y_1^2 – 4y_1 + 4) = (x_1^2 – 6x_1 + 9)+(y_1^2 + 8y_1 + 16)$$

$$\Rightarrow 4x_1 – 12y_1 – 20 = 0$$

$$\Rightarrow 4(x_1 – 3y_1 – 5) = 0$$

$$\Rightarrow x_1 – 3y_1 – 5 = 0$$

Also, $$SA = SC$$

$$\Rightarrow SA^2 = SC^2$$

$$\Rightarrow (x_1 – 1)^2 + (y_1 – 2)^2 = (x_1 – 5)^2 + (y_1 + 6)^2$$

$$\Rightarrow (x_1^2 – 2x_1 + 1) + (y_1^2 – 4y_1 + 4) = (x_1^2 – 10x_1 + 25) + (y_1^2 + 12y_1 + 36)$$

$$\Rightarrow 8x_1 – 16y_1 – 56 = 0$$

$$\Rightarrow 8(x_1 – 2y_1 – 7) = 0$$

$$\Rightarrow x_1 – 2y_1 – 7 = 0$$

(1) – (2) $$\Rightarrow -y_1 + 2 = 0$$

$$\Rightarrow y_1 = 2$$

(1) $$\Rightarrow x_1 – 6 – 5 = 0$$

$$\Rightarrow x_1 = 11$$

Centre of the circle $$S(x_1,y_1) = (11,2)$$

Also, we have $$A = (1,2)$$

So, radius $$r = SA$$

$$\Rightarrow r^2 = SA^2$$

$$r^2 = (11-1)^2 + (2-2)^2 = 10^2 = 100$$

Circle with centre (11,2) and $$r^2 = 100$$

$$(x-11)^2+(y-2)^2 = 100$$

$$\Rightarrow x^2 + y^2 – 22x – 4y + 25 = 0$$

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LAQ-2 : Find the equation of the circle passing through the points (3,4), (3,2) and (1,4)

Let $$A = (3,4) B = (3,2) C = (1,4)$$

We take $$S(x_1,y_1)$$ as the centre of the circle

$$\Rightarrow SA = SB = SC$$

Now $$SA = SB$$

$$\Rightarrow SA^2 = SB^2$$

$$\Rightarrow (x_1 – 3)^2 + (y_1 – 4)^2 = (x_1 – 3)^2 + (y_1 – 2)^2$$

$$\Rightarrow (y_1 – 4)^2 = (y_1 – 2)^2$$

$$\Rightarrow (y_1 – 4) = \pm(y_1 – 2)$$

$$\Rightarrow (y_1 – 4) = \pm(y_1 – 2)$$

$$\Rightarrow (y_1 – 4) = (y_1 – 2)$$ (or) $$(y_1 – 4) = -(y_1 – 2)$$

Now $$y_1 – 4 = -(y_1 – 2) = -y_1 + 2$$

$$\Rightarrow 2y_1 = 6$$

$$\Rightarrow y_1 = 3$$

Also $$SA = SC$$

$$\Rightarrow SA^2 = SC^2$$

$$\Rightarrow (x_1 – 3)^2 + (y_1 – 4)^2 = (x_1 – 1)^2 + (y_1 – 4)^2$$

$$\Rightarrow (x_1 – 3)^2 = (x_1 – 1)^2$$

$$\Rightarrow (x_1 – 3) = \pm(x_1 – 1)$$

$$\Rightarrow (x_1 – 3) = (x_1 – 1)$$ (or) $$(x_1 – 3) = -(x_1 – 1)$$

Now $$x_1 – 3 = -(x_1 – 1) = -x_1 + 1$$

$$\Rightarrow 2x_1 = 4$$

$$\Rightarrow x_1 = 2$$

From (1) & (2) we get

Centre of the circle $$S(x_1,y_1) = (2,3)$$

Also, we have $$A = (3,4)$$

So, radius $$r = SA$$

$$\Rightarrow r^2 = SA^2$$

$$r^2 = (2-3)^2 + (3-4)^2 = 1 + 1 = 2$$

Circle with centre (2,3) and $$r^2 = 2$$

$$(x-2)^2 + (y-3)^2 = 2$$

$$\Rightarrow x^2 + y^2 – 4x – 6y + 11 = 0$$

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LAQ-3 : Find the equation of circle passing through the points (2,1), (5,5), (-6,7)

Let $$A = (2,1) B = (5,5) C = (-6,7)$$

Let $$S(x_1, y_1)$$ be the centre of the circle

$$\Rightarrow SA = SB = SC$$

Now, $$SA = SB$$

$$\Rightarrow SA^2 = SB^2$$

$$\Rightarrow (x_1 – 2)^2 + (y_1 – 1)^2 = (x_1 – 5)^2 + (y_1 – 5)^2$$

$$\Rightarrow x_1^2 + 4 – 4x_1 + y_1^2 + 1 – 2y_1 = x_1^2 + 25 – 10x_1 + y_1^2 + 25 – 10y_1$$

$$\Rightarrow -4x_1 – 2y_1 + 5 = -10x_1 – 10y_1 + 50$$

$$\Rightarrow 6x_1 + 8y_1 – 45 = 0$$

Also, $$SA = SC$$

$$\Rightarrow SA^2 = SC^2$$

$$\Rightarrow (x_1 – 2)^2 + (y_1 – 1)^2 = (x_1 + 6)^2 + (y_1 – 7)^2$$

$$\Rightarrow x_1^2 + 4 – 4x_1 + y_1^2 + 1 – 2y_1 = x_1^2 + 36 + 12x_1 + y_1^2 + 49 – 14y_1$$

$$\Rightarrow -4x_1 – 2y_1 + 5 = 12x_1 – 14y_1 + 85$$

$$\Rightarrow 16x_1 – 12y_1 + 80 = 0$$

$$\Rightarrow 4x_1 – 3y_1 + 20 = 0$$

Solving (1) & (2) we get the centre $$S(x_1,y_1)$$

$$x_1 = -\frac{1}{2} y_1 = 6$$

Centre of the circle $$S(x_1,y_1) = (-\frac{1}{2},6)$$

Also, we have $$A = (2,1)$$

So, radius $$r = SA$$

$$\Rightarrow r^2 = SA^2$$

$$r^2 = \left(2 + \frac{1}{2}\right)^2 + (1 – 6)^2 = \left(\frac{5}{2}\right)^2 + (-5)^2 = \frac{25}{4} + 25 = \frac{125}{4}$$

Equation of the circle with centre $$(-1/2,6)$$ and $$r^2 = \frac{125}{4}$$

$$(x + \frac{1}{2})^2 + (y – 6)^2 = \frac{125}{4}$$

$$\Rightarrow x^2 + y^2 + x – 12y + 5 = 0$$

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LAQ-4 : Show that the points (1,1), (-6,0), (-2,2) and (-2,-8) are concyclic

Let $$A = (1,1) B = (-6,0) C = (-2,2) D = (-2,-8)$$

We take $$S(x_1,y_1)$$ as the centre of the circle

$$\Rightarrow SA = SB = SC$$

Now, $$SA = SB$$

$$\Rightarrow SA^2 = SB^2$$

$$\Rightarrow (x_1-1)^2+(y_1-1)^2 = (x_1+6)^2+(y_1-0)^2$$

$$\Rightarrow (x_1^2 – 2x_1 + 1) + (y_1^2 – 2y_1 + 1) = (x_1^2 + 12x_1 + 36) + y_1^2$$

$$\Rightarrow 14x_1 + 2y_1 + 34 = 0$$

$$\Rightarrow 7x_1 + y_1 + 17=0$$

Also, $$SB = SC$$

$$\Rightarrow SB^2 = SC^2$$

$$\Rightarrow (x_1+6)^2 + (y_1-0)^2 = (x_1+2)^2 + (y_1-2)^2$$

$$\Rightarrow (x_1^2 + 12x_1 + 36) + y_1^2 = (x_1^2 + 4x_1 + 4) + (y_1^2 – 4y_1 + 4)$$

$$\Rightarrow 8x_1 + 4y_1 + 28 = 0$$

$$\Rightarrow 2x_1 + y_1 + 7 = 0$$

Solving (1) & (2) we get the centre $$S(x_1,y_1)$$

From the equations:

$$\Rightarrow 5x_1 + 10 = 0$$

$$\Rightarrow x_1 = -2$$

From (1):

$$\Rightarrow 7(-2) + y_1 + 17 = 0$$

$$\Rightarrow y_1 + 3 = 0$$

$$\Rightarrow y_1 = -3$$

Centre of the circle is $$S(x_1,y_1) = (-2,-3)$$

Also, we have $$A = (1,1)$$

$$\Rightarrow r^2 = SA^2$$

$$r^2 = (1+2)^2 + (1+3)^2 = 9 + 16 = 25$$

Equation of the circle with centre (−2,−3) and $$r^2 = 25$$

$$(x+2)^2 + (y+3)^2 = 25$$

$$\Rightarrow x^2 + y^2 + 4x + 6y – 12 = 0$$

Now putting D(−2,−8) in the above equation then we get

$$(-2)^2 + (-8)^2 + 4(-2) + 6(-8) – 12 = 4 + 64 – 8 – 48 – 12 = 68 – 68 = 0$$

D(-2,-8) lies on the circle

The given 4 points are concyclic

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LAQ-5 : If (2,0), (0,1), (4,5), (0,c) are concyclic then find c

Let $$A = (2,0) B = (0,1) C = (4,5) D = (0,c)$$

We take $$S(x_1,y_1)$$ as the centre of the circle

$$\Rightarrow SA = SB = SC$$

Now, $$SA = SB$$

$$\Rightarrow SA^2 = SB^2$$

$$\Rightarrow (x_1-2)^2+(y_1-0)^2 = (x_1-0)^2 + (y_1-1)^2$$

$$\Rightarrow (x_1^2 – 4x_1 + 4) + (y_1^2) = (x_1^2) + (y_1^2 – 2y_1 + 1)$$

$$\Rightarrow 4x_1 – 2y_1 + 1 – 4 = 0$$

$$\Rightarrow 4x_1 – 2y_1 – 3 = 0$$

Also, $$SB = SC$$

$$\Rightarrow SB^2 = SC^2$$

$$\Rightarrow (x_1 – 0)^2 + (y_1 – 1)^2 = (x_1 – 4)^2 + (y_1 – 5)^2$$

$$\Rightarrow (x_1^2) + (y_1^2 – 2y_1 + 1) = (x_1^2 – 8x_1 + 16) + (y_1^2 – 10y_1 + 25)$$

$$\Rightarrow 8x_1 – 2y_1 + 10y_1 + 1 – 16 – 25 = 0$$

$$\Rightarrow 8x_1 + 8y_1 – 40 = 0$$

$$\Rightarrow 8(x_1 + y_1 – 5) = 0$$

$$\Rightarrow x_1 + y_1 – 5 = 0$$

Solving (1) & (2) we get the centre $$S(x_1,y_1)$$

2× Equation (2) $$2x_1 + 2y_1 – 10 = 0$$

Adding Equation (1) to Equation (3) gives:

$$6x_1 = 13$$

$$\Rightarrow x_1 = \frac{13}{6}$$

From Equation (2):

$$y_1 = 5 – x_1 = 5 – \frac{13}{6} = \frac{30-13}{6} = \frac{17}{6}$$

Centre of the circle is $$S(x_1,y_1) = \left(\frac{13}{6}, \frac{17}{6}\right)$$

Also, we have $$A = (2,0)$$

Radius $$r = SA$$

$$\Rightarrow r^2 = (2-\frac{13}{6})^2 + (0-\frac{17}{6})^2 = \frac{1}{36} + \frac{289}{36} = \frac{290}{36}$$

Circle with centre $$\left(\frac{13}{6},\frac{17}{6}\right)$$ and $$r^2 = \frac{290}{36}$$

$$\left(x-\frac{13}{6}\right)^2 + \left(y-\frac{17}{6}\right)^2 = \frac{290}{36}$$

For D(0,c) in the circle equation

$$\left(0-\frac{13}{6}\right)^2 + \left(c-\frac{17}{6}\right)^2 = \frac{290}{36}$$

$$\Rightarrow \left(c-\frac{17}{6}\right)^2 = \frac{121}{36}$$

$$\Rightarrow 6c-17 = \pm11$$

For $$6c – 17 = 11$$

$$6c = 28$$

$$\Rightarrow c = \frac{28}{6} = \frac{14}{3}$$

For $$6c – 17 = -11$$

$$6c = 6$$

$$\Rightarrow c = 1$$

Thus, $$c = \frac{14}{3}$$ or $$c = 1$$

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LAQ-6 : Find the equation of the circle passing through (4,1) (6,5) and having the centre on the line 4x + 3y -24 = 0

Let $$A = (4,1) B = (6,5)$$

We take $$S(x_1,y_1)$$ as the centre of the circle

$$\Rightarrow SA = SB$$

$$\Rightarrow SA^2 = SB^2$$

$$\Rightarrow (x_1-4)^2 + (y_1-1)^2 = (x_1-6)^2 + (y_1-5)^2$$

$$\Rightarrow (x_1^2 – 8x_1 + 16) + (y_1^2 – 2y_1 + 1) = (x_1^2 – 12x_1 + 36) + (y_1^2 – 10y_1 + 25)$$

$$\Rightarrow 17 – 8x_1 – 2y_1 = 61 – 12x_1 – 10y_1$$

$$\Rightarrow 4x_1 + 8y_1 – 44 = 0$$

But the centre $$(x_1,y_1)$$ lies on $$4x + 3y – 24 = 0$$

$$\Rightarrow 4x_1 + 3y_1 – 24 = 0$$

Subtracting (1) from (2) gives:

$$\Rightarrow -5y_1 + 20 = 0$$

$$\Rightarrow 5y_1 = 20$$

$$\Rightarrow y_1 = 4$$

From (2), $$4x_1 + 3(4) – 24 = 0$$

$$\Rightarrow 4x_1 + 12 – 24 = 0$$

$$\Rightarrow 4x_1 – 12 = 0$$

$$\Rightarrow 4x_1 = 12$$

$$\Rightarrow x_1 = 3$$

Centre of the circle $$S(x_1,y_1) = (3,4)$$

Also, we have $$A = (4,1)$$

So, radius $$r = SA$$

$$\Rightarrow r^2 = (3-4)^2 + (4-1)^2 = 1 + 9 = 10$$

Circle with centre (3,4) and $$r^2 = 10$$

$$(x-3)^2+(y-4)^2 = 10$$

$$\Rightarrow x^2 + y^2 – 6x – 8y + 15 = 0$$

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LAQ-7 : Find the equation of the circle, which passes through (2,-3), (-4,5) and whose centre lies on the line 4x + 3y + 1 = 0

Let $$A = (2,-3) B = (-4,5)$$

We take $$S(x_1,y_1)$$ as the centre of the circle

$$\Rightarrow SA = SB$$

$$\Rightarrow SA^2 = SB^2$$

$$\Rightarrow (x_1-2)^2 + (y_1+3)^2 = (x_1+4)^2 + (y_1-5)^2$$

$$\Rightarrow (x_1^2 + 4x_1 + 16) + (y_1^2 + 6y_1 + 9) = (x_1^2 + 8x_1 + 16) + (y_1^2 – 10y_1 + 25)$$

$$\Rightarrow 12x_1 – 16y_1 + 28 = 0$$

$$\Rightarrow 4(3x_1 – 4y_1 + 7) = 0$$

$$\Rightarrow 3x_1 – 4y_1 + 7 = 0$$

But centre $$(x_1,y_1)$$ lies on $$4x + 3y + 1 = 0$$

$$\Rightarrow 4x_1 + 3y_1 + 1 = 0$$

Solving (1) & (2) we get the centre $$S(x_1,y_1)$$

$$x_1 = -1 y_1 = 1$$

$$S(x_1,y_1) = (-1,1)$$

Also, we have $$A = (2,-3)$$

So, radius $$r = SA$$

$$\Rightarrow r^2 = SA^2$$

$$\Rightarrow r^2 = (-1-2)^2 + (1+3)^2 = 9 + 16 = 25$$

Circle with centre (−1,1) and $$r^2 = 25$$

$$(x+1)^2+(y-1)^2 = 25$$

$$\Rightarrow (x^2+2x+1)+(y^2-2y+1)=25$$

$$\Rightarrow x^2+y^2+2x-2y-23=0$$

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LAQ-8 : Show that the circles x²+y²-6x-2y+1=0 and x²+y²+2x-8y+13=0 touch each other. Find the point of contact and the equation of the common tangent at their point of contact

First circle is $$S = x^2 + y^2 – 6x – 2y + 1 = 0$$

Centre $$C_1 = (3,1)$$ radius $$r_1 = \sqrt{9+1-1} = 3$$

Second circle is $$S = x^2 + y^2 + 2x – 8y + 13 = 0$$

Centre $$C_2 = (-1,4)$$ radius $$r_2 = \sqrt{1+16-13} = 2$$

$$C_1C_2 = \sqrt{(3+1)^2+(1-4)^2} = \sqrt{16+9} = \sqrt{25} = 5$$

Also $$r_1 + r_2 = 3 + 2 = 5 = C_1C_2$$

The circles touch each other externally

Now $$r_1 : r_2 = 3 : 2$$

So the point of contact P divides the join of $$C_1(3,1) C_2(-1,4)$$ internally in the ratio $$3:2$$

$$P = \left(\frac{3(1)+2(3)}{5},\frac{3(4)+2(1)}{5}\right) = \left(\frac{3}{5},\frac{14}{5}\right)$$

The equation of the common tangent to the circles $$S = 0$$ and $$S’ = 0$$ at the point of contact is given by $$S-S’ = 0$$

$$\Rightarrow (x^2+y^2-6x-2y+1)-(x^2+y^2+2x-8y+13)=0$$

$$\Rightarrow -8x+6y-12=0$$

$$\Rightarrow 4x-3y+6=0$$

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LAQ-9 : Show that the circles x²+y²-6x-9y+13=0, x²+y²-2x-16y=0 touch each other. Find the point of contact and the equation of the common tangent at that point

First circle is $$S = x^2 + y^2 – 6x – 9y + 13 = 0$$

Centre $$C_1 = (3,\frac{9}{2})$$ radius $$r_1 = \sqrt{9 + \left(\frac{9}{2}\right)^2 – 13} = \sqrt{9 + \frac{81}{4} – 13} = \sqrt{\frac{36+81-52}{4}} = \sqrt{\frac{65}{2}}$$

Second circle is $$S’ = x^2 + y^2 – 2x – 16y = 0$$

Centre $$C_2 = (1,8)$$ radius $$r_2 = \sqrt{1^2 + 8^2 – 0} = \sqrt{1 + 64} = \sqrt{65}$$

$$C_1C_2 = \sqrt{(3-1)^2 + \left(\frac{9}{2} – 8\right)^2} = \sqrt{4 + \frac{1}{4}} = \sqrt{\frac{16+1}{4}} = \sqrt{\frac{65}{4}} = \sqrt{\frac{65}{2}}$$

Also $$r_2 – r_1 = \sqrt{65} – \sqrt{\frac{65}{2}} = \sqrt{\frac{65}{2}} = C_1C_2$$

The circles touch each other internally

Now $$r_1 : r_2 = \sqrt{\frac{65}{2}} : \sqrt{65} = \frac{1}{2} : 1 = 1 : 2$$

So, the point of contact P divides the join of $$C_1(3,\frac{9}{2}) C_2(1,8)$$ externally in the ratio $$1:2$$

$$P = \left(\frac{1(1)-2(3)}{1-2}, \frac{1(8)-2(\frac{9}{2})}{1-2}\right) = (-5,-1)$$

The equation of the common tangent to the circles $$S = 0$$ and $$S’ = 0$$ at the point of contact is given by $$S-S’ = 0$$

$$\Rightarrow (x^2+y^2-6x-9y+13)-(x^2+y^2-2x-16y)=0$$

$$\Rightarrow -4x+7y+13=0$$

$$\Rightarrow 4x-7y-13=0$$

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LAQ-10 : Find the equation of direct common tangents to the circles x²+y²+22x-4y-100=0, x²+y²-22x+4y+100=0

For the circle $$x^2 + y^2 + 22x – 4y – 100 = 0$$ Centre $$C_1 = (-11,2)$$ Radius $$r_1 = \sqrt{121+4+100} = \sqrt{225} = 15$$

For the circle $$x^2 + y^2 – 22x + 4y + 100 = 0$$ Centre $$C_2 = (11,-2)$$ Radius $$r_2 = \sqrt{121+4+100} = \sqrt{25} = 5$$

Now $$C_1C_2 = \sqrt{(-11-11)^2 + (2+2)^2} = \sqrt{484+16} = \sqrt{500} = 10\sqrt{5}$$

$$r_1+r_2 = 15+5 = 20$$

Hence $$C_1C_2 > r_1+r_2$$

Also $$r_1:r_2 = 15:5 = 3:1$$

External centre of similitude E divides C₁​C₂​ in the ratio 3:1 externally

$$\Rightarrow E = (3(11)-(1)(-11)/3-1, 3(-2)-1(2)/3-1) = (22,-4)$$

Method I : Tangent through 22,−4 with slope ‘m’ is

$$y+4=m(x-22)$$

$$\Rightarrow mx-y-22m-4 = 0….(1)$$

The perpendicular distance from C₂​(11,−2) to (1) is r₂ ​= 5

$$\Rightarrow |11m+2-22m-4|/\sqrt{m^2+1} = 5$$

$$\Rightarrow |11m+2| = 5\sqrt{m^2+1}$$

$$\Rightarrow (11m+2)^2 = 25(m^2+1)$$

$$\Rightarrow 121m^2+4+44m = 25m^2+25$$

$$\Rightarrow 96m^2+44m-21=0$$

$$\Rightarrow 96m^2+72m-28m-21=0$$

$$\Rightarrow 24m(4m+3)-7(4m+3) = 0$$

$$\Rightarrow (24m-7)(4m+3)=0$$

$$\Rightarrow m = 7/24 m = -3/4$$

Tangent through 22,−4 with slope 7/24 is

$$\Rightarrow 24(y+4)=7(x-22)$$

$$\Rightarrow 7x-24y-178=0$$

Tangent through 22,−4 with slope −3/4 is

$$\Rightarrow 4(y+4)=-3(x-22)$$

$$\Rightarrow 3x+4y-50=0$$

Method II : The pair of direct common tangents through A(22,−4) to the circle

$$S = x^2 + y^2 + 22x 4y – 100 = 0$$ is $$S_{12} = S_{11}(S)$$

$$\Rightarrow [22x-4y+11(x+22)-2(y-4)-100]^2 = (22^2 + 4^2 + 22(22)-4(-4)-100)(x^2+y^2+22x-4y-100)$$

$$\Rightarrow (33x-6y+150)^2 = 900(x^2+y^2+22x-4y-100)$$

$$\Rightarrow 9(11x-2y+50)^2 = 900(x^2+y^2+22x-4y-100)$$

$$\Rightarrow (11x-2y+50)^2 = 100(x^2+y^2+22x-4y-100)$$

$$\Rightarrow 121x^2+4y^2+2500-44xy-200y+1100x = 100x^2+100y^2+2200x-400y-10000$$

$$\Rightarrow 21x^2-96y^2-44xy+1100x+200y+12500=0$$

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LAQ-11 : Find the equation to the pair of transverse common tangents to the circles x²+y²-4x-10y+28=0 and x²+y²+4x-6y+4=0

For the circle $$x^2 + y^2 – 4x – 10y + 28 = 0$$ Centre $$C_1 = (2,5)$$ radius $$r_1 = \sqrt{(-2)^2+(-5)^2-28} = \sqrt{1} = 1$$

For the circle $$x^2 + y^2 + 4x – 6y + 4 = 0$$ Centre $$C_2 = (-2,3)$$ radius $$r_2 = \sqrt{2^2+(-3)^2-4} = \sqrt{9} = 3$$

The internal centre of similitude I divides C_1​C_2​ internally in the ratio $$r_1 : r_2 = 1 : 3$$

$$I = \left(\frac{1(-2)+3(2)}{1+3}, \frac{1(3)+3(5)}{1+3}\right) = \left(\frac{4}{4}, \frac{18}{4}\right) = (1, \frac{9}{2})$$

The equation to the pair of transverse common tangents is $$S_{12} = S_{1} – S_{2}$$

$$\Rightarrow [x^2 + y^2 – 4x – 10y + 28] – [x^2 + y^2 + 4x – 6y + 4] = 0$$

$$\Rightarrow -4x – 10y + 28 – 4x + 6y – 4 = 0$$

$$\Rightarrow -8x – 4y + 24 = 0$$

$$3x^2 + 4xy – 24x – 4y + 21 = 0$$

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LAQ-12 : Find all common tangents to the circles x²+y²-2x-6y+6=0,x²+y²=1

For the circle $$x^2 + y^2 – 2x – 6y + 6 = 0$$ Centre $$C_1 = (1,3)$$ radius $$r_1 = \sqrt{(-1)^2 + (-3)^2 – 6} = \sqrt{4} = 2$$

For the circle $$x^2 + y^2 = 1$$ Centre $$C_2 = (0,0)$$ radius $$r_2 = 1$$

Now $$C_1C_2 = \sqrt{1^2+3^2} = \sqrt{10}$$

Also $$r_1+r_2 = 1+2 = 3$$

Here $$C_1C_2 > r_1+r_2$$

The two circles are such that one lies entirely outside the other

The internal centre of the similitude I divides C_1​C₂​ internally in the ratio 2:1

$$\Rightarrow I = \left(\frac{2(0)+1(1)}{2+1}, \frac{2(0)+1(3)}{2+1}\right) = \left(\frac{1}{3}, 1\right)$$

The external centre of the similitude E divides C₁​C₂​ externally in the ratio 2:1

$$\Rightarrow E = \left(\frac{2(0)-1(1)}{2-1}, \frac{2(0)-1(3)}{2-1}\right) = (-1, -3)$$

Equation of transverse common tangents drawn through $$I\left(\frac{1}{3},1\right)$$ to $$S = x^2 + y^2 – 1 = 0$$ is $$S_{11}S=S_{12}$$

$$\Rightarrow \left(\frac{1}{9}+1-1\right)(x^2+y^2-1)=\left(\frac{x}{3}+y-1\right)^2$$

$$\Rightarrow \frac{1}{9}(x^2 + y^2 – 1) = \frac{1}{9}(x + 3y – 3)^2$$

$$\Rightarrow x^2 + y^2 – 1 = x^2 + 9y^2 + 9 + 6xy – 18y – 6x$$

$$\Rightarrow 8y^2 + 6xy – 6x – 18y + 10 = 0$$

$$\Rightarrow 2(4y^2 + 3xy – 3x – 9y + 5) = 0$$

$$\Rightarrow 4y^2 + 3xy – 3x – 9y + 5 = 0$$

Equation of direct common tangents drawn through $$E(-1,3)$$ to $$S = x^2 + y^2 – 1 = 0$$ is $$S_{11}S=S_{12}$$

$$\Rightarrow (1+9-1)(x^2+y^2-1) = (x+3y+1)^2$$

$$\Rightarrow 9(x^2+y^2-1) = x^2+9y^2+1+6xy+6y+2x$$

$$\Rightarrow 8x^2 – 6xy – x – 3y – 5 = 0$$