System Of Circles (VSAQs)

Maths-2B | 2. System Of Circles – VSAQs:
Welcome to VSAQs in Chapter 2: System Of Circles. This page contains the most important VSAQs in this chapter. Aim to secure top marks in your exams by understanding these clear and straightforward Very Short Answer Questions.

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VSAQ-1 : Find K if the pair of circles x²+y²-5x-14y-34=0, x²+y²+2x+4y+k=0 are orthogonal

From the given circles, we get $$g = -\frac{5}{2} f = -7 c = -34$$ and $$g’ = 1 f’ = 2 c’ = k$$

Using the orthogonal condition $$2gg’ + 2ff’ = c + c’$$

$$\Rightarrow 2\left(-\frac{5}{2}\right)(1) + 2(-7)(2) = -34 + k$$

$$\Rightarrow -5 – 28 = -34 + k$$

$$\Rightarrow k = -33 + 34 = 1$$

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VSAQ-2 : Find K if the pairs of circles x²+y²+4x+8=0,x²+y²-16y+k=0 are orthogonal

From the given circles, we have $$g = 2 f = 0 c = 8$$ and $$g’ = 0 f’ = -8 c’ = k$$

Using the orthogonal condition $$2gg’ + 2ff’ = c + c’$$

$$\Rightarrow 2(2)(0) + 2(0)(-8) = 8 + k$$

$$\Rightarrow 0 + 0 = 8 + k$$

$$\Rightarrow k = -8$$

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VSAQ-3 : S.T the circles x²+y²-2x-2y-7=0, 3x²+3y²-8x+29y=0 intersect each other orthogonally

Here $$g = -1 f = -1 c = -7$$ and $$g’ = -\frac{4}{3} f’ = \frac{29}{6} c’ = 0$$

Using the orthogonal condition $$2gg’ + 2ff’ = c + c’$$

L.H.S $$= 2gg’ + 2ff’$$

$$= 2(-1)\left(-\frac{4}{3}\right) + 2(-1)\left(\frac{29}{6}\right)$$

$$= \frac{8}{3} – \frac{29}{3} = -\frac{21}{3} = -7$$

R.H.S $$= c + c’ = -7 + 0 = -7$$

L.H.S = R.H.S,

Hence, the 2 circles cut orthogonally.

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VSAQ-4 : Show that the circles x²+y²-2lx+g=0,x²+y²+2my-g=0 intersect each other orthogonally

Here $$g = -l f = 0 c = g$$ and $$g’ = 0 f’ = m c’ = -g$$

Using the orthogonal condition $$2gg’ + 2ff’ = c + c’$$

$$\text{L.H.S} = 2(-l)(0) + 2(0)(m) = 0 + 0 = 0$$

$$\text{R.H.S} = c + c’ = g – g = 0$$

$$\text{L.H.S} = \text{R.H.S}$$

Hence, the two circles cut orthogonally.

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VSAQ-5 : Find the equation of the radical axis of the circles 2x²+2y²+3x+6y-5=0,3x²+3y²-7x+8y-11=0

Given circles are

$$S = x^2 + y^2 + \frac{3}{2}x + 3y – \frac{5}{2} = 0$$ and $$S’ = x^2 + y^2 – \frac{7}{3}x + \frac{8}{3}y – \frac{11}{3} = 0$$

Radical axis is $$S – S’ = 0$$

$$\Rightarrow \left(\frac{3}{2} + \frac{7}{3}\right)x + \left(3 – \frac{8}{3}\right)y + \left(-\frac{5}{2} + \frac{11}{3}\right) = 0$$

$$\Rightarrow \left(\frac{9}{6} + \frac{14}{6}\right)x + \left(\frac{9}{3} – \frac{8}{3}\right)y + \left(-\frac{15}{6} + \frac{22}{6}\right) = 0$$

$$\Rightarrow \frac{23}{6}x + \frac{1}{3}y + \frac{7}{6} = 0$$

$$\Rightarrow 23x + 2y + 7 = 0$$

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VSAQ-6 : Find the equation of the radical axis of x²+y²+4x+6y-7=0, 4(x²+y²)+8x+12y-9=0

Let $$S = x^2 + y^2 + 4x + 6y – 7 = 0$$ and $$S’ = x^2 + y^2 + 2x + 3y – \frac{9}{4} = 0$$

Radical axis is $$S – S’ = 0$$

$$\Rightarrow (4-2)x + (6-3)y + (-7+\frac{9}{4}) = 0$$

$$\Rightarrow 2x + 3y – \frac{19}{4} = 0$$

$$\Rightarrow 8x + 12y – 19 = 0$$

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VSAQ-7 : Show that the angle between the circles x²+y²=a²,x²+y²=ax+ay is 3π/4

Given circles are $$x^2 + y^2 – a^2 = 0$$

$$x^2 + y^2 – ax – ay = 0$$

Centre $$C1 = (0,0) C2 = \left(\frac{a}{2}, \frac{a}{2}\right)$$

Distance $$d = C1C2 = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{a}{2}\right)^2} = \sqrt{\frac{2a^2}{4}} = \frac{a}{\sqrt{2}}$$

Also, from (1), $$r1 = a$$

From (2), $$r2 = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{a}{2}\right)^2} = \sqrt{\frac{a^2}{2}} = \frac{a}{\sqrt{2}}$$

Cosine of angle between the circles (θ) is given by:

$$\cos \theta = \frac{d^2 – r1^2 – r2^2}{2r1r2}$$

Substituting the values:

$$\cos \theta = \frac{\left(\frac{a^2}{2}\right) – a^2 – \left(\frac{a^2}{2}\right)}{2 \cdot a \cdot \frac{a}{\sqrt{2}}} = \frac{-a^2}{2a^2/\sqrt{2}} = -\frac{1}{\sqrt{2}} = \cos 135^\circ$$

Thus, the angle between the circles is $$\theta = 135^\circ = \frac{3\pi}{4}$$

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VSAQ-8 : Find the angle between the circles x²+y²-12x-6y+41=0, x²+y²+4x+6y-59=0

Given circles are

$$x^2 + y^2 – 12x – 6y + 41 = 0$$

$$x^2 + y^2 + 4x + 6y – 59 = 0$$

Centre $$C1 = (6,3) C2 = (-2,-3)$$

Distance $$d = C1C2 = \sqrt{(6+2)^2 + (3+3)^2}$$

$$= \sqrt{64 + 36} = \sqrt{100} = 10$$

From (1), $$r_1 = \sqrt{(-6)^2 + (-3)^2 – 41}$$

$$= \sqrt{36 + 9 – 41}$$

$$= \sqrt{4} = 2$$

From (2), $$r_2 = \sqrt{2^2 + 3^2 + 59}$$

$$= \sqrt{4 + 9 + 59}$$

$$= \sqrt{72}$$

Cosine of the angle between the circles (θ) is given by:

$$\cos \theta = \frac{d^2 – r_1^2 – r_2^2}{2r_1r_2}$$

$$= \frac{100 – 4 – 72}{2 \times 2 \times \sqrt{72}}$$

$$= \frac{24}{4 \times 6 \sqrt{2}}$$

$$= \frac{24}{24\sqrt{2}}$$

$$= \frac{1}{\sqrt{2}} = \cos 45^\circ$$

Thus, the angle between the circles is $$\theta = 45^\circ = \frac{\pi}{4}$$

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VSAQ-9 : If the angle between the circles x²+y²-12x-6y+41=0,x²+y²+kx+6y-59=0 is 45° then find k

From the given circles we get $$g = -6 f = -3 c = 41$$ and $$g’ = \frac{k}{2} f’ = 3 c’ = 59$$

If θ is the angle between the given circles then

$$\cos \theta = \frac{(c + c’) – (2gg’ + 2ff’)}{2\sqrt{(g^2 + f^2 – c)(g’^2+f’^2 – c’)}}$$

$$\Rightarrow \cos 45^\circ = \frac{(41 + 59) – 2(-6)\left(\frac{k}{2}\right) – 2(-3)(3)}{2\sqrt{36 + 9 – 41} \sqrt{\left(\frac{k^2}{4}\right) + 9 – 59}}$$

$$\Rightarrow \frac{1}{\sqrt{2}} = \frac{100 – 6k – 18}{2\sqrt{4} \sqrt{\frac{k^2}{4} + 68}}$$

$$\Rightarrow 1/\sqrt{2} = \frac{6k}{4\sqrt{\frac{k^2}{4} + 68}} = \frac{3k}{2\sqrt{\frac{k^2}{4} + 68}}$$

$$\Rightarrow 1/\sqrt{2} = \frac{3k}{\sqrt{k^2 + 272}}$$

On squaring and cross multiplying, we get

$$k^2 + 272 = 2(9k^2) = 18k^2$$

$$\Rightarrow 17k^2 = 272$$

$$\Rightarrow k^2 = \frac{272}{17} = 16$$

$$\Rightarrow k = \pm 4$$