9 Most VSAQ’s of System of Circle Chapter in Inter 2nd Year Maths-2B (TS/AP)

2 Marks

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

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

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

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

$$\Rightarrow k = \pm 4$$