Circle (SAQs)

Maths-2B | 1. Circle – SAQs:
Welcome to SAQs in Chapter 1: Circle. This page contains the most Important FAQs for Short 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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SAQ-1 : Find the length of the chord intercepted by the circle x²+y²-8x-2y-8=0 on the line x+y+1=0

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

It’s Centre $$C = (4,1)$$ radius $$r = \sqrt{16 + 1 + 8} = \sqrt{25} = 5$$

Perpendicular distance from the centre (4,1) to the line $$x + y + 1 = 0$$

$$P = \frac{|4 + 1 + 1|}{\sqrt{1^2 + 1^2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}$$

Length of the chord

$$= 2\sqrt{r^2 – p^2} = 2\sqrt{5^2 – (3\sqrt{2})^2}$$

$$= 2\sqrt{25 – 18} = 2\sqrt{7}$$

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SAQ-2 : Find the length of the chord intercepted by the circle x²+y²-x+3y-22=0 on the line y=x-3

Given circle $$x^2 + y^2 – x + 3y – 22 = 0$$

Its Center $$C = \left(\frac{1}{2}, -\frac{3}{2}\right)$$ radius $$r = \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{3}{2}\right)^2 + 22} = \sqrt{\frac{1}{4} + \frac{9}{4} + 22}$$

$$= \sqrt{\frac{1}{4} + \frac{9}{4} + \frac{88}{4}} = \sqrt{\frac{98}{4}} = \sqrt{\frac{49}{2}} = \frac{7}{\sqrt{2}}$$

Perpendicular distance from the centre $$\left(\frac{1}{2},-\frac{3}{2}\right)$$ to the line $$y = x – 3$$

$$⇒ x – y – 3 = 0$$

$$P = \frac{| \frac{1}{2} – \left(-\frac{3}{2}\right) – 3|}{\sqrt{1^2 + (-1)^2}} = \frac{| \frac{1}{2} + \frac{3}{2} – 3|}{\sqrt{2}} = \frac{| 1 – 3 |}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$$

Length of the chord

$$= 2\sqrt{r^2 – p^2} = 2\sqrt{\left(\frac{7}{\sqrt{2}}\right)^2 – \left(\frac{1}{\sqrt{2}}\right)^2} = 2\sqrt{\frac{49}{2} – \frac{1}{2}} = 2\sqrt{\frac{48}{2}} = 2\sqrt{24}$$

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SAQ-3 : Find the mid point of the chord intercepted by x²+y²-2x-10y+1=0 on the line x – 2y + 7 =0. Also find the length of the chord

(i) To find the length of the chord:

The perpendicular distance from $$C(1,5)$$ to $$x – 2y + 7 = 0$$ is $$P = \left| \frac{1 – 10 + 7}{\sqrt{1^2 + (-2)^2}} \right| = \frac{2}{\sqrt{5}}$$

Length of the chord is $$2\sqrt{r^2 – p^2} = 2\sqrt{25 – \frac{4}{5}} = 2\sqrt{\frac{125 – 4}{5}} = 2\sqrt{\frac{121}{5}} = \frac{2 \times 11}{\sqrt{5}} = \frac{22}{\sqrt{5}}$$

(ii) To find the midpoint of the chord:

For the given circle $$x^2 + y^2 – 2x – 10y + 1 = 0$$ centre $$C = (-g,-f) = (1,5)$$ radius $$r = \sqrt{(-1)^2 + (-5)^2 – 1} = 5$$

Midpoint of the chord = Foot of the perpendicular (h,k) from the centre C=(1,5) onto

$$x – 2y + 7 = 0$$

$$\begin{align} \frac{h-1}{1} &= \frac{k-5}{-2} = \frac{-(1 – 10 + 7)}{1^2 + (-2)^2} \ \frac{h-1}{1} &= \frac{k-5}{-2} = \frac{2}{5} \ \end{align}$$

Thus, $$(\frac{7}{5}, \frac{21}{5})$$ is the midpoint of the given chord.

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SAQ-4 : If a point P is moving such that the lengths of tangents drawn from P to the circles x²+y²-4x-6y-12=0 and x²+y²+6x+18y+26=0 are in the ratio 2:3 then find the equation of the locus of P

Let $$P(x_1, y_1)$$ be a point on the locus.

Let PT_1​ be the length of the tangent from P to $$S = x^2 + y^2 – 4x – 6y – 12 = 0$$

and PT₂​ be the length of the tangent from P to $$S’ = x^2 + y^2 + 6x + 18y + 26 = 0$$

Given that $$PT_1 : PT_2 = 2 : 3$$

$$\Rightarrow \frac{PT_1}{PT_2} = \frac{2}{3}$$

$$\Rightarrow 3PT_1 = 2PT_2$$

$$\Rightarrow 9(PT_1)^2 = 4(PT_2)^2$$

$$\Rightarrow 9[x_1^2 + y_1^2 – 4x_1 – 6y_1 – 12] = 4[x_1^2 + y_1^2 + 6x_1 + 18y_1 + 26]$$

$$\Rightarrow 9x_1^2 + 9y_1^2 – 36x_1 – 54y_1 – 108 = 4x_1^2 + 4y_1^2 + 24x_1 + 72y_1 + 104$$

$$\Rightarrow 5x_1^2 + 5y_1^2 – 60x_1 – 126y_1 – 212 = 0$$

Equation of the locus of $$P(x_1, y_1)$$ is $$5x^2 + 5y^2 – 60x – 126y – 212 = 0$$

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SAQ-5 : Show that x + y + 1 = 0 is touches the circle x²+y²-3x+7y+14=0 and find its point of contact

(i) To show that x + y + 1 = 0 touches the circle:

For the given circle $$x^2 + y^2 – 3x + 7y + 14 = 0$$ Centre $$C = \left(\frac{3}{2},-\frac{7}{2}\right)$$ radius $$r = \sqrt{g^2 + f^2 – c} = \sqrt{\left(\frac{3}{2}\right)^2 + \left(-\frac{7}{2}\right)^2 – 14} = \sqrt{\frac{9}{4} + \frac{49}{4} – 14} = \sqrt{\frac{2}{4}} = \frac{1}{\sqrt{2}}$$

The perpendicular distance from $$C\left(\frac{3}{2},-\frac{7}{2}\right)$$ to $$x + y + 1 = 0$$

$$P = \frac{\left|\frac{3}{2} + \left(-\frac{7}{2}\right) + 1\right|}{\sqrt{1^2 + 1^2}} = \frac{\left|-\frac{4}{2}\right|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$$

Since $$P = r$$ the line $$x + y + 1 = 0$$ touches the circle.

(ii) To find the point of contact:

Point of contact = foot of the perpendicular (h,k) from $$C\left(\frac{3}{2},-\frac{7}{2}\right)$$ onto $$x + y + 1 = 0$$

$$h – x_1/a = k – y_1/b = -(ax_1 + by_1 + c)/(a^2 + b^2)$$

$$h – \frac{3}{2}/1 = k + \frac{7}{2}/1 = -\left(\frac{3}{2} + \left(-\frac{7}{2}\right) + 1\right)/(1^2 + 1^2) = \frac{1}{2}$$

Thus, the point of contact is $$(h,k) = (2, -3)$$

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SAQ-6 : Find the equation of tangent of x²+y²-2x+4y=0 at (3,-1). Also find the equation of other tangent parallel to it

(i) To find the tangent at (3,−1):

The equation of the tangent at (3,−1) on $$S = x^2 + y^2 – 2x + 4y = 0$$ is $$S_1 = 0$$

$$\Rightarrow x(3) + y(-1) – (x + 3) + 2(y – 1) = 0$$

$$\Rightarrow 3x – y – x – 3 + 2y – 2 = 0$$

$$\Rightarrow 2x + y – 5 = 0$$

(ii) To find the parallel tangent:

Slope of the tangent $$2x + y – 5 = 0$$ is $$m = -2$$

Also $$x^2 + y^2 – 2x + 4y = 0$$

$$\Rightarrow g = -1 f = 2$$

Radius $$r = \sqrt{1^2 + 2^2 – 0} = \sqrt{5}$$

Formula: Equation of the tangents with slope m is $$y + f = m(x + g) \pm r\sqrt{1 + m^2}$$

$$\Rightarrow (y + 2) = -2(x – 1) \pm \sqrt{5} \sqrt{1 + 4}$$

$$\Rightarrow (y + 2) = -2(x – 1) \pm 5$$

$$\Rightarrow 2x + y \pm 5 = 0$$

The equation of the parallel tangent is $$2x + y + 5 = 0$$

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SAQ-7 : Find the equations of the tangents to the circle x²+y²-4x+6y-12=0 and parallel to the line x + y – 8 =0

For the given circle $$x^2 + y^2 – 4x + 6y – 12 = 0$$ the centre is $$C = (2,3)$$ and the radius $$r = \sqrt{(-2)^2 + (3)^2 + 12} = \sqrt{4 + 9 + 12} = \sqrt{25} = 5$$

The given line is $$x + y – 8 = 0$$

Any line parallel to the given line is of the form $$x + y + k = 0$$

$$\Rightarrow \left| \frac{2 + 3 + k}{\sqrt{1^2 + 1^2}} \right| = 5$$

$$\Rightarrow \left| \frac{k + 5}{\sqrt{2}} \right| = 5$$

$$\Rightarrow |k + 5| = 5\sqrt{2}$$

$$\Rightarrow k + 5 = \pm 5\sqrt{2}$$

$$\Rightarrow k = -5 \pm 5\sqrt{2}$$

Hence, the equations of the required tangents are: $$x + y + 1 \pm 5\sqrt{2} = 0$$

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SAQ-8 : Find the pole of the line x + y + 2 = 0 w.r.t the circle x²+y²-4x+6y-12=0

Let $$P(x_1,y_1)$$ be the pole of $$x + y + 2 = 0$$

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

$$S = x^2 + y^2 – 4x + 6y – 12 = 0$$ is $$S_1 = 0$$

$$\Rightarrow x_1x + y_1y – 2(x_1 + x) + 3(y_1 + y) – 12 = 0$$

$$\Rightarrow (x_1 – 2)x + (y_1 + 3)y – 2x_1 + 3y_1 – 12 = 0$$

Comparing the above with x + y + 2 = 0 we get

$$\frac{x_1 – 2}{1} = \frac{y_1 + 3}{1} = \frac{-2x_1 + 3y_1 – 12}{2} = k$$

$$\Rightarrow x_1 – 2 = k; \quad y_1 + 3 = k; \quad \text{and} \quad -2x_1 + 3y_1 – 12 = 2k$$

$$\Rightarrow x_1 = k + 2, \quad y_1 = k – 3$$

Solving $$-2(k + 2) + 3(k – 3) – 12 = 2k$$

$$\Rightarrow -2k – 4 + 3k – 9 – 12 = 2k$$

$$\Rightarrow k = -25$$

$$x_1 = k + 2 = -25 + 2 = -23 \quad \text{and} \quad y_1 = k – 3 = -25 – 3 = -28$$

Thus, the pole $$P(x_1,y_1) = (-23,-28)$$

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SAQ-9 : Find the pole of the line 3x + 4y – 45 = 0 w.r.t the circle x²+y²-6x-8y+5=0

Let $$P(x_1,y_1)$$ be the pole of $$3x + 4y – 45 = 0$$

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

$$S = x^2 + y^2 – 6x – 8y + 5 = 0$$ is $$S_1 = 0$$

$$\Rightarrow x_1x + y_1y – 3(x_1 + x) – 4(y_1 + y) + 5 = 0$$

$$\Rightarrow (x_1 – 3)x + (y_1 – 4)y – 3x_1 – 4y_1 + 5 = 0$$

Comparing the above with $$3x + 4y – 45 = 0$$ we get

$$\frac{x_1 – 3}{3} = \frac{y_1 – 4}{4} = \frac{-3x_1 – 4y_1 + 5}{-45} = k$$

$$\Rightarrow x_1 – 3 = 3k; \quad y_1 – 4 = 4k; \quad \text{and} \quad -3x_1 – 4y_1 + 5 = -45k$$

Solving for x₁​ and y₁​: $$x_1 = 3 + 3k; \quad y_1 = 4 + 4k$$

Substituting x₁​ and y_1​ into $$3x_1 + 4y_1 – 5 = 45k$$

$$3(3 + 3k) + 4(4 + 4k) – 5 = 45k$$

$$9 + 9k + 16 + 16k – 5 = 45k$$

$$25 + 25k – 5 = 45k$$

$$20 + 25k = 45k$$

$$20k = 20$$

$$k = 1$$

Therefore, $$x_1 = 3 + 3(1) = 6$$ and $$y_1 = 4 + 4(1) = 8$$

Pole $$P(x_1,y_1) = (6,8)$$

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SAQ-10 : Find the value of K if Kx + 3y – 1 = 0 and 2x + y + 5 = 0 are conjugate with respect to the circle x²+y²-2x-4y-4=0

From the given lines, we have $$l_1 = k m_1 = 3 n_1 = -1 l_2 = 2 m_2 = 1 n_2 = 5$$

The given circle is $$x^2 + y^2 – 2x – 4y – 4 = 0$$

$$\Rightarrow g = -1 f = -2 c = -4$$

radius $$r = \sqrt{(-1)^2 + (-2)^2 + 4} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$$

Conjugate lines condition: $$r^2(l_1l_2 + m_1m_2) = (l_1g + m_1f – n_1)(l_2g + m_2f – n_2)$$

$$\Rightarrow 3^2[(k \cdot 2) + 3 \cdot 1] = [k(-1) + 3(-2) + 1][2(-1) + 1(-2) – 5]$$

$$\Rightarrow 9(2k + 3) = (-k – 6 + 1)(-2 – 2 – 5)$$

$$\Rightarrow 9(2k + 3) = (-k – 5)(-9)$$

$$\Rightarrow 18k + 27 = 9k + 45$$

$$\Rightarrow 18k – 9k = 45 – 27$$

$$\Rightarrow 9k = 18$$

$$\Rightarrow k = 2$$

Hence, the value of k is 2.