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

SAQ-1 : Find the radical centre of the circles x²+y²-4x-6y+5=0,x²+y²-2x-4y-1=0,x²+y²-6x-2y=0

The given circles are

$$S = x^2 + y^2 – 4x – 6y + 5 = 0$$

$$S’ = x^2 + y^2 – 2x – 4y – 1 = 0$$

$$S” = x^2 + y^2 – 6x – 2y = 0$$

One radical axis is $$S – S’ = 0$$

$$\Rightarrow (-4x + 2x) + (-6y + 4y) + (5 + 1) = 0$$

$$\Rightarrow -2x – 2y + 6 = 0$$

$$\Rightarrow x + y – 3 = 0 \quad \text{(1)}$$

Another radical axis is $$S – S” = 0$$

$$\Rightarrow (-4x + 6x) + (-6y + 2y) + 5 = 0$$

$$\Rightarrow 2x – 4y + 5 = 0 \quad \text{(2)}$$

Now, multiply (1) by 2:

$$\Rightarrow 2x + 2y – 6 = 0 \quad \text{(3)}$$

Subtract (2) from (3):

$$\Rightarrow 6y – 11 = 0$$

$$\Rightarrow y = \frac{11}{6}$$

From (1), solve for x:

$$x = 3 – y = 3 – \frac{11}{6} = \frac{18}{6} – \frac{11}{6} = \frac{7}{6}$$

The radical centre is $$\left(\frac{7}{6}, \frac{11}{6}\right)$$

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SAQ-2 : Find the radical centre of circles x²+y²+4x-7=0, 2x²+2y²+3x+5y-9=0, x²+y²+y=0

The given circles are

$$S = x^2 + y^2 + 4x – 7 = 0$$

$$\Rightarrow 2x^2 + 2y^2 + 8x – 14 = 0$$

$$S’ = 2x^2 + 2y^2 + 3x + 5y – 9 = 0$$

$$S” = x^2 + y^2 + y = 0$$

$$\Rightarrow 2×2 + 2y2 + 2y = 0$$

One radical axis is $$S – S’ = 0$$

$$\Rightarrow (8x – 3x) – 5y – 14 + 9 = 0$$

$$\Rightarrow 5x – 5y – 5 = 0$$

$$\Rightarrow 5(x – y – 1) = 0$$

$$\Rightarrow x – y – 1 = 0$$

Another radical axis is $$S – S” = 0$$

$$\Rightarrow 8x – 2y – 14 = 0$$

$$\Rightarrow 2(4x – y – 7) = 0$$

$$\Rightarrow 4x – y – 7 = 0$$

(2) – (1)

$$\Rightarrow 3x – 6 = 0$$

$$\Rightarrow x = 2$$

From (1) $$\Rightarrow 2 – y – 1 = 0$$

$$\Rightarrow y = 1$$

The radical centre is $$(x,y) = (2,1)$$

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SAQ-3 : Find the equation and length of the common chord of the two circles x²+y²+2x+2y+1=0,x²+y²+4x+3y+2=0

Given circles are $$S = x^2 + y^2 + 2x + 2y + 1 = 0$$ and

$$S’ = x^2 + y^2 + 4x + 3y + 2 = 0$$

Equation of the common chord is $$S-S’ = 0$$

$$\Rightarrow -2x – y – 1 = 0$$

$$\Rightarrow 2x + y + 1 = 0$$

For the circle $$S = x^2 + y^2 + 2x + 2y + 1 = 0$$ Center $$C (-1,-1)$$

radius $$r = \sqrt{1^2 + 1^2 – 1} = \sqrt{1 + 1 – 1} = \sqrt{1} = 1$$

Length of the perpendicular from $$C(-1,-1)$$ to the line $$2x + y + 1 = 0$$ is

$$P = \frac{| 2(-1) – 1 + 1 |}{\sqrt{2^2 + 1^2}} = \frac{|-2|}{\sqrt{5}} = \frac{2}{\sqrt{5}}$$

Length of the common chord

$$= 2\sqrt{r^2 – p^2} = 2\sqrt{1 – \left(\frac{4}{5}\right)} = 2\sqrt{\frac{1}{5}} = \frac{2}{\sqrt{5}} \text{ units}$$

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SAQ-4 : Find the equation and length of the common chord of the two circles. x²+y²+3x+5y+4=0, x²+y²+5x+3y+4=0

Given circles are $$S = x^2 + y^2 + 3x + 5y + 4 = 0$$ and $$S’ = x^2 + y^2 + 5x + 3y + 4 = 0$$

Equation of the common chord is $$S – S’ = 0$$

$$\Rightarrow 3x – 5x + 5y – 3y + 4 – 4 = 0$$

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

$$\Rightarrow x – y = 0$$

For the circle $$S = x^2 + y^2 + 3x + 5y + 4 = 0$$ centre $$C(-3/2,-5/2)$$

$$r = \sqrt{\left(\frac{9}{4}\right) + \left(\frac{25}{4}\right) – 4} = \sqrt{\frac{9+25-16}{4}} = \sqrt{\frac{18}{4}} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}}$$

Length of the perpendicular from $$C(-3/2,-5/2)$$ to the line $$x – y = 0$$ is $$P = \frac{|-3/2 + 5/2|}{\sqrt{1^2 + (-1)^2}} = \frac{1}{\sqrt{2}}$$

Length of the common chord

$$= 2\sqrt{r^2 – p^2} = 2\sqrt{\left(\frac{9}{2}\right)-\left(\frac{1}{2}\right)} = 2\sqrt{\frac{8}{2}} = 2\cdot 2 = 4 \text{ units}$$

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SAQ-5 : If x + y = 3 is the equation of the chord AB of the circle x²+y²-2x+4y-8=0, find the equation of the circle having (AB) ̅ as diameter

Given Circle is $$S = x^2 + y^2 – 2x + 4y – 8 = 0$$ and the given line is $$L = x + y – 3 = 0$$

Equation of any circle passing through the points of intersections of

$$S = 0 L = 0 S + \lambda L = 0$$

$$\Rightarrow (x^2 + y^2 – 2x + 4y – 8) + \lambda(x + y – 3) = 0$$

$$\Rightarrow x^2 + y^2 – (2 – \lambda)x + (4 + \lambda)y – (3\lambda + 8) = 0 \quad …(1)$$

Centre of the above circle is $$\left(\frac{2-\lambda}{2}, \frac{4+\lambda}{2}\right)$$

Line L = 0 becomes a diameter if the above centre lies on $$L = x + y – 3 = 0$$

$$\Rightarrow \left(\frac{2-\lambda}{2}\right) + \left(\frac{4+\lambda}{2}\right) – 3 = 0$$

$$\Rightarrow 2 – \lambda + 4 + \lambda – 6 = 0$$

$$\Rightarrow 2\lambda = -8$$

$$\Rightarrow \lambda = -4$$

From (1) required circle’s equation is

$$x^2 + y^2 – (2 + 4)x + (4 – 4)y – (-12 + 8) = 0$$

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

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

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SAQ-6 : If the straight line 2x + 3y = 1 intersects the circle x²+y²=4 at the points A and B, find the equation of the circle having AB as diameter

Given circle is $$S = x^2 + y^2 – 4 = 0$$ and the given line is $$L = 2x + 3y – 1 = 0$$

Equation of any circle passing through the points of intersections of $$S = 0 L = 0 S + \lambda L = 0$$

$$\Rightarrow (x^2 + y^2 – 4) + \lambda(2x + 3y – 1) = 0$$

$$\Rightarrow x^2 + y^2 + 2\lambda x + 3\lambda y – 4 – \lambda = 0 \quad …(1)$$

Centre of the above circle is $$(-λ, -\frac{3\lambda}{2})$$

Line L = 0 becomes a diameter if the above centre lies on $$L = 2x + 3y – 1 = 0$$

$$\Rightarrow 2(-\lambda) + 3\left(-\frac{3\lambda}{2}\right) – 1 = 0$$

$$\Rightarrow -2\lambda – \frac{9\lambda}{2} – 1 = 0$$

$$\Rightarrow -\frac{4\lambda}{2} – \frac{9\lambda}{2} = 1$$

$$\Rightarrow -\frac{13\lambda}{2} = 1$$

$$\Rightarrow 13\lambda = -2$$

$$\Rightarrow \lambda = -\frac{2}{13}$$

From (1) the required circle’s equation is

$$x^2 + y^2 – \frac{4}{13}x – \frac{6}{13}y – 4 + \frac{2}{13} = 0$$

$$\Rightarrow 13x^2 + 13y^2 – 4x – 6y – 50 = 0$$

Therefore, the equation of the required circle is $$13x^2 + 13y^2 – 4x – 6y – 50 = 0$$

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SAQ-7 : Find the equation of the circle passing through the points of intersection of the circles x²+y²-8x-6y+21=0,x²+y²-2x-15=0 and (1,2)

Given circles are $$S = x^2 + y^2 – 8x – 6y + 21 = 0$$ $$S’ = x^2 + y^2 – 2x – 15 = 0$$

Radical axis of the circles is $$L = S – S’ = 0$$

$$\Rightarrow -8x + 2x – 6y + 21 + 15 = 0$$

$$\Rightarrow -6x – 6y + 36 = 0$$

$$\Rightarrow x + y – 6 = 0$$

Equation of any circle passing through the points of intersection of $$S’ = 0 L = 0 S’ + \lambda L$$

$$\Rightarrow (x^2 + y^2 – 2x – 15) + \lambda(x + y – 6) = 0 \quad…(1)$$

(1) passes through the point (1,2)

$$\Rightarrow (1 + 4 – 2 – 15) + \lambda(1 + 2 – 6) = 0$$

$$\Rightarrow -12 + \lambda(-3) = 0$$

$$\Rightarrow 3\lambda = -12$$

$$\Rightarrow \lambda = -4$$

Put λ=−4 in (1) then $$(x^2 + y^2 – 2x – 15) – 4(x + y – 6) = 0$$

$$\Rightarrow x^2 + y^2 – 2x – 15 – 4x – 4y + 24 = 0$$

$$\Rightarrow x^2 + y^2 – 6x – 4y + 9 = 0$$

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SAQ-8 : Find the equation of the circle which passes through the origin and intersects the circles x²+y²-4x+6y+10=0 and x²+y²+12y+6=0 orthogonally

Given that the equation of the required circle is $$S = x^2 + y^2 + 2gx + 2fy + c = 0$$

If the circle S = 0 passes through the origin O(0,0) then $$0^2 + 0^2 + 2g(0) + 2f(0) + c = 0$$

$$\Rightarrow c = 0$$

Given $$S = 0$$ is orthogonal to $$x^2 + y^2 – 4x + 6y + 10 = 0$$

Using the condition for orthogonality between two circles, $$2gg’ + 2ff’ = c + c’$$

$$\Rightarrow 2g(-2) + 2f(3) = 0 + 10$$

$$\Rightarrow -4g + 6f = 10$$

Given S = 0 is also orthogonal to $$x^2 + y^2 + 12y + 6 = 0$$

Again, using the orthogonality condition,

$$\Rightarrow 2g(0) + 2f(6) = 0 + 6$$

$$\Rightarrow 12f = 6$$

$$\Rightarrow f = \frac{1}{2}$$

From (2), using $$f = \frac{1}{2}$$

$$-4g + 6(\frac{1}{2}) = 10$$

$$-4g + 3 = 10$$

$$\Rightarrow -4g = 7$$

$$\Rightarrow g = -\frac{7}{4}$$

Substituting the values of g, f, and c in, we get the equation of the circle as:

$$x^2 + y^2 – \frac{7}{2}x + y = 0$$

Or equivalently,

$$2x^2 + 2y^2 – 7x + 2y = 0$$

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SAQ-9 : Find the equation of the circle passing through the points (0,-3) and cutting the circles x²+y²-6x+3y+5=0,x²+y²-x-7y=0 orthogonally

Given the equation of the required circle is $$S = x^2 + y^2 + 2gx + 2fy + c = 0$$

S = 0 passes through the point (0,−3)

$$\Rightarrow 0 + 9 + 2g(0) + 2f(-3) + c = 0$$

$$\Rightarrow 6f = 9 + c \quad…(1)$$

S = 0 is orthogonal to $$x^2 + y^2 – 6x + 3y + 5 = 0$$

$$\Rightarrow 2g(-3) + 2f(3/2) = c + 5$$

$$\Rightarrow -6g + 3f = c + 5 \quad…(2)$$

S = 0 is orthogonal to $$x^2 + y^2 – x – 7y = 0$$

$$\Rightarrow 2g(-1/2) + 2f(-7/2) = c + 0$$

$$\Rightarrow g + 7f = -c \quad…(3)$$

From (2) and (3) we get

$$-5g + 10f = 5$$

$$\Rightarrow -5(g-2f) = 5$$

$$\Rightarrow g-2f = -1 \quad…(4)$$

From (3) and (1) we get

$$g + 13f = 9 \quad…(5)$$

From (4) and (5) we find f and g, and then c

$$f = 2/3, g = 1/3, c = -5$$

Substituting the values of g, f, c in, we get the equation of the required circle as

$$x^2 + y^2 + 2(1/3)x + 2(2/3)y – 5 = 0$$

$$\Rightarrow 3x^2 + 3y^2 + 2x + 4y – 15 = 0$$