Circle (VSAQs)

Maths-2B | 1. Circle – VSAQs:
Welcome to VSAQs in Chapter 1: Circle. 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 : If the circle x²+y²-4x+6y+a=0 has radius 4 then find a

Given circle is $$x^2 + y^2 – 4x + 6y + a = 0$$

$$\Rightarrow g = -2 f = 3 c = a$$

But radius $$r = 4$$

$$\sqrt{g^2 + f^2 – c} = 4$$

$$\Rightarrow \sqrt{(-2)^2 + 3^2 – a} = 4$$

$$\Rightarrow \sqrt{4 + 9 – a} = 4$$

$$\Rightarrow \sqrt{13 – a} = 4$$

On squaring both sides:

$$13 – a = 16$$

$$\Rightarrow a = 13 – 16 = -3$$

Hence, the value of a is −3.

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VSAQ-2 : If x²+y²-4x+6y+c=0 represents a circle with radius 6, find the value of c

Given circle is $$x^2 + y^2 – 4x + 6y + c = 0$$

$$\Rightarrow g = -2 f = 3 c = c$$

But radius r = 6

$$\sqrt{g^2 + f^2 – c} = 6$$

$$\Rightarrow \sqrt{(-2)^2 + 3^2 – c} = 6$$

$$\Rightarrow \sqrt{4 + 9 – c} = 6$$

$$\Rightarrow \sqrt{13 – c} = 6$$

On squaring both sides:

$$13 – c = 36$$

$$\Rightarrow c = 13 – 36 = -23$$

Hence, the value of c is −23.

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VSAQ-3 : If x²+y²+2gx+2fy=0 has centre (-4,-3) then find g, f and radius

Given the circle equation is $$x^2 + y^2 + 2gx + 2fy = 0$$

But the centre $$C = (-4,-3)$$

Given $$(-g,-f) = (-4,-3)$$

We find that $$g = 4$$ and $$f = 3$$

Thus, the circle equation becomes $$x^2 + y^2 + 4x + 6y = 0$$

The radius r is calculated as $$\sqrt{g^2 + f^2 – c}$$

$$r = \sqrt{(4)^2 + (3)^2 – 0} = \sqrt{16 + 9} = \sqrt{25} = 5$$

Therefore, the radius of the given circle is 5.

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VSAQ-4 : If x²+y²+2gx+2fy-12=0 is a circle with centre (2,3) then find (g,f) and its radius

Given circle is $$x^2 + y^2 – 4x – 6y – 12 = 0$$

But Centre $$C = (2,3)$$

$$\Rightarrow (-g,-f) = (2,3)$$

$$\Rightarrow g = -2 f = -3$$

Circle is $$x^2 + y^2 – 4x – 6y – 12 = 0$$

Radius $$r = \sqrt{g^2 + f^2 – c}$$

$$= \sqrt{(-2)^2 + (-3)^2 + 12}$$

$$= \sqrt{4 + 9 + 12} = \sqrt{25} = 5$$

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VSAQ-5 : Find value of ‘a’ if 2x²+ay²-3x+2y-1=0 represents a circle and also find its radius

In the equation of a circle, we know Coefficient of x² = Coefficient of y²

$$\Rightarrow 2 = a$$

Circle is $$2x^2 + 2y^2 – 3x + 2y – 1 = 0$$

$$\Rightarrow x^2 + y^2 – \frac{3}{2}x + y – \frac{1}{2} = 0$$

$$\Rightarrow g = -\frac{3}{4}, f = \frac{1}{2}, c = -\frac{1}{2}$$

Radius $$r = \sqrt{\left(\frac{3}{4}\right)^2 + \left(\frac{1}{2}\right)^2 + \frac{1}{2}}$$

$$\Rightarrow r = \sqrt{\frac{9}{16} + \frac{1}{4} + \frac{1}{2}} = \sqrt{\frac{9+4+8}{16}} = \sqrt{\frac{21}{4}}$$

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VSAQ-6 : Find the values of a,b if ax²+bxy+3y²-5x+2y-3=0 represents a circle. Also find the radius and centre

In the equation of a circle, we have:

(i) Coefficient of x² = Coefficient of y² $$\Rightarrow a = 3$$

(ii) Coefficient of $$xy = 0 \Rightarrow b = 0$$

Circle is $$3x^2 + 3y^2 – 5x + 2y – 3 = 0$$

$$\Rightarrow x^2 + y^2 – \frac{5}{3}x + \frac{2}{3}y – 1 = 0$$

$$\Rightarrow g = -\frac{5}{6} f = \frac{1}{3} c = -1$$

Centre $$C = \left(\frac{5}{6}, -\frac{1}{3}\right)$$

$$r = \sqrt{\left(\frac{5}{6}\right)^2 + \left(\frac{1}{3}\right)^2 + 1} = \sqrt{\frac{25}{36} + \frac{1}{9} + 1} = \sqrt{\frac{25}{36} + \frac{4}{36} + \frac{36}{36}} = \sqrt{\frac{65}{36}} = \frac{\sqrt{65}}{6}$$

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VSAQ-7 : Find the parametric equations of the circle x²+y²-6x+4y-12=0

Given circle is $$x^2 + y^2 – 6x + 4y – 12 = 0$$

Centre $$C = (-g,-f) = (3,-2)$$

Radius $$r = \sqrt{(-3)^2 + (2)^2 + 12} = \sqrt{9 + 4 + 12} = \sqrt{25} = 5$$

Parametric equations:

$$x = -g + r\cos\theta y = -f + r\sin\theta$$

$$\Rightarrow x = 3 + 5\cos\theta y = -2 + 5\sin\theta$$

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VSAQ-8 : Find the Parametric equations of the circle x²+y²+6x+8y-96=0

Given circle is $$x^2 + y^2 + 6x + 8y – 96 = 0$$

Centre $$C = (-g,-f) = (-3,-4)$$ radius $$r = \sqrt{(-3)^2 + (-4)^2 + 96}$$

$$= \sqrt{9 + 16 + 96} = \sqrt{121} = 11$$

Parametric equations:

$$x = -g + r\cos\theta y = -f + r\sin\theta$$

$$\Rightarrow x = -3 + 11\cos\theta y = -4 + 11\sin\theta$$

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VSAQ-9 : Obtain the parametric equations of the circle (x-3)²+(y-4)²=8²

Given circle is $$(x-3)^2+(y-4)^2=8^2$$

Centre $$C = (-g,-f) = (3,4)$$

$$\Rightarrow g = -3 f = -4$$ radius $$r = 8$$

Parametric equations:

$$x = g + r\cos\theta y = f + r\sin\theta$$

$$\Rightarrow x = 3 + 8\cos\theta y = 4 + 8\sin\theta$$

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VSAQ-10 : Obtain the parametric equations of the circle 4(x²+y²)=9

Given circle is $$4(x^2 + y^2) = 9$$

$$\Rightarrow x^2 + y^2 = \frac{9}{4} = \left(\frac{3}{2}\right)^2$$

Centre $$C = (-g,-f) = (0,0)$$

$$\Rightarrow g = 0 f = 0$$ radius $$r = \frac{3}{2}$$

Parametric equations:

$$x = r\cos\theta y = r\sin\theta$$

$$\Rightarrow x = \left(\frac{3}{2}\right)\cos\theta y = \left(\frac{3}{2}\right)\sin\theta$$

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VSAQ-11 : Find the equation of the circle passing through (-2,3) and with centre at (0,0)

Given centre $$C = (0,0)$$

Point on the circle $$P(-2,3)$$

radius $$r = CP = \sqrt{(0+2)^2 + (0-3)^2}$$

$$= \sqrt{4 + 9} = \sqrt{13}$$

Equation of the circle is $$x^2 + y^2 = 13$$

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VSAQ-12 : Find the equation of the circle whose centre is (-1,2) and which passes through (5,6)

Given Centre $$C = (1,2)$$ point on the circle $$P(5,6)$$

radius $$r = CP = \sqrt{(1-5)^2 + (2-6)^2} = \sqrt{(-4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32}$$

Circle with centre (1,2) and radius $$\sqrt{32}$$ ​ is

$$(x-1)^2 + (y-2)^2 = 32$$

$$\Rightarrow (x^2 – 2x + 1) + (y^2 – 4y + 4) = 32$$

$$\Rightarrow x^2 + y^2 – 2x – 4y – 27 = 0$$

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VSAQ-13 : Find the equation of the circle passing through the point (-2,14) and concentric with x²+y²-6x-4y-12=0

Equation of the required concentric circle is $$x^2 + y^2 – 6x – 4y + k = 0$$

(1) passes through $$P(-2,14)$$

$$\Rightarrow (-2)^2 + 14^2 – 6(-2) – 4(14) + k = 0$$

$$\Rightarrow 4 + 196 + 12 – 56 + k = 0$$

$$\Rightarrow k = -156$$

From (1) the required concentric circle is $$x^2 + y^2 – 6x – 4y – 156 = 0$$

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VSAQ-14 : Find the equation of the circle passing through the point (2,3) and concentric with x²+y²+8x+12y+15=0

Equation of the required concentric circle is $$x^2 + y^2 + 8x + 12y + k = 0$$

Since (1) passes through $$P(2,3)$$

$$\Rightarrow 2^2 + 3^2 + 8(2) + 12(3) + k = 0$$

$$\Rightarrow 4 + 9 + 16 + 36 + k = 0$$

$$\Rightarrow 65 + k = 0$$

$$\Rightarrow k = -65$$

Therefore, from (1) the required concentric circle is $$x^2 + y^2 + 8x + 12y – 65 = 0$$

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VSAQ-15 : Find the value of K if the length of the tangent from (5,4) to x²+y²+2ky=0 is 1

Length of the tangent from (5,4) to

$$S = x^2 + y^2 + 2ky = 0$$ is $$\sqrt{S_{11}} = 1$$

On squaring both sides we get $$S_{11} = 1$$

$$\Rightarrow 5^2 + 4^2 + 2k(4) = 1$$

$$\Rightarrow 25 + 16 + 8k = 1$$

$$\Rightarrow 41 + 8k = 1$$

$$\Rightarrow 8k = -40$$

$$\Rightarrow k = -40/8 = -5$$

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VSAQ-16 : Find the value of K if the length of the tangent from (2,5) to x²+y²-5x+4y+k=0 is √37

Length of the tangent from (2,5) to $$S = x^2 + y^2 – 5x + 4y + k = 0$$ is $$\sqrt{S_{11}} = \sqrt{37}$$

On squaring both sides we get $$S_{11} = 37$$

$$\Rightarrow (2)^2 + (5)^2 – 5(2) + 4(5) + k = 37$$

$$\Rightarrow 4 + 25 – 10 + 20 + k = 37$$

$$\Rightarrow 39 + k = 37$$

$$\Rightarrow k = -2$$

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VSAQ-17 : Find the power of the point P(-1,1) w.r.to the circle x²+y²-6x+4y-12=0

Power of $$P(x_1,y_1) = (-1,1)$$ with respect to

$$S = x^2 + y^2 – 6x + 4y – 12$$ is $$S_{11}$$

$$= x_1^2 + y_1^2 – 6x_1 + 4y_1 – 12$$

$$= (-1)^2 + 1^2 – 6(-1) + 4(1) – 12$$

$$= 1 + 1 + 6 + 4 – 12 = 0$$

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VSAQ-18 : Find the length of the tangent from (1,3) to the circle x²+y²-2x+4y-11=0

Length of tangent from (1,3) to the circle

$$S = x^2 + y^2 – 2x + 4y – 11 = 0$$ is $$\sqrt{S_{11}}$$

$$= \sqrt{1^2 + 3^2 – 2(1) + 4(3) – 11} = \sqrt{1 + 9 – 2 + 12 – 11} = \sqrt{9} = 3$$

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VSAQ-19 : Find the value of K if the points (1,3),(2,k) are conjugate w.r.to the circle x²+y²=35

The points (1,3), (2,k) are conjugate to w.r.t the circle $$S = x^2 + y^2 – 35 = 0$$

$$\Rightarrow S_{12} = 0$$

$$\Rightarrow x_1x_2 + y_1y_2 – 35 = 0$$

$$\Rightarrow (1)(2) + (3)(k) – 35 = 0$$

$$\Rightarrow 2 + 3k = 35$$

$$\Rightarrow 3k = 33$$

$$\Rightarrow k = 11$$

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VSAQ-20 : Find the value of K if the points (4,2), (K,-3) are conjugate with respect to the circle x²+y²-5x+8y+6=0

The points (4,2), (k,-3) are conjugate w.r.t the circle $$S = x^2 + y^2 – 5x + 8y + 6 = 0$$

$$\Rightarrow S_{12} = 0$$

$$\Rightarrow 4k + 2(-3) – \frac{5}{2}(4+k) + 4(2-3) + 6 = 0$$

$$\Rightarrow 4k – 6 – 10 – \frac{5k}{2} – 4 + 6 = 0$$

$$\Rightarrow 4k – \frac{5k}{2} – 14 = 0$$

$$\Rightarrow \frac{3k}{2} = 14$$

$$\Rightarrow k = \frac{28}{3}$$

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VSAQ-21 : Find the pole of ax+by+c=0 with respect to x²+y²=r²

Let $$P(x_1,y_1)$$ be the pole.

The equation of the polar of $$P(x_1,y_1)$$ with respect to

$$S = 0 S_1 = 0$$

$$\Rightarrow x_1x + y_1y – r^2 = 0$$

Comparing with $$ax + by + c = 0$$ we get

$$x_1/a = y_1/b = -r^2/c$$

$$\Rightarrow x_1 = -ar^2/c, \quad y_1 = -br^2/c$$

Thus, Pole $$P(x_1,y_1) = (-ar^2/c,-br^2/c)$$

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VSAQ-22 : Find the equation of the polar of (1,-2) w.r.to x²+y²-10x-10y+25=0

Polar of P(1,2) w.r.t the circle

$$S = x^2 + y^2 – 10x – 10y + 25 = 0$$ is $$S_1 = 0$$

$$\Rightarrow x_1x + y_1y + g(x_1+x) + f(y_1+y) + c = 0$$

$$\Rightarrow 1(x) + 2(y) – 5(1+x) – 5(2+y) + 25 = 0$$

$$\Rightarrow x + 2y – 5 – 5x – 10 – 5y + 25 = 0$$

$$\Rightarrow -4x – 7y + 30 = 0$$

$$4x + 7y – 30 = 0$$