RBSE Solutions Class 11 Maths Chapter 12 Conic Section Exercise 12.2

Question 1. Find the point of intersection of circle \(x^2 + y^2 = 25\) and line \(4x + 3y = 12\) and also find length of intersecting chord.
Answer:
Given the equation of the circle: \(x^2 + y^2 = 25\) ...(i)
And the equation of the line: \(4x + 3y = 12\) ...(ii)
From equation (ii), we can express \(y\) in terms of \(x\):
\(3y = 12 - 4x\)
\( \implies y = \frac{12 - 4x}{3} \)
\( \implies y = 4 - \frac{4}{3}x \) ...(iii)
Now, substitute the value of \(y\) from equation (iii) into equation (i):
\(x^2 + \left(4 - \frac{4}{3}x\right)^2 = 25\)
\( \implies x^2 + 16 - 2 \cdot 4 \cdot \frac{4}{3}x + \frac{16}{9}x^2 = 25\)
\( \implies x^2 + 16 - \frac{32}{3}x + \frac{16}{9}x^2 = 25\)
Multiply the entire equation by 9 to remove fractions:
\(9x^2 + 144 - 9 \cdot \frac{32}{3}x + 16x^2 = 225\)
\( \implies 9x^2 + 144 - 96x + 16x^2 = 225\)
\( \implies 25x^2 - 96x + 144 - 225 = 0\)
\( \implies 25x^2 - 96x - 81 = 0\)
This is a quadratic equation in the form \(ax^2 + bx + c = 0\), where \(a=25\), \(b=-96\), and \(c=-81\).
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\(x = \frac{-(-96) \pm \sqrt{(-96)^2 - 4(25)(-81)}}{2(25)}\)
\( \implies x = \frac{96 \pm \sqrt{9216 + 8100}}{50}\)
\( \implies x = \frac{96 \pm \sqrt{17316}}{50}\)
\( \implies x = \frac{96 \pm 6\sqrt{481}}{50}\)
\( \implies x = \frac{48 \pm 3\sqrt{481}}{25}\)
Now, substitute these values of \(x\) back into equation (iii) to find \(y\):
\(y = 4 - \frac{4}{3}x\)
\( \implies y = 4 - \frac{4}{3} \left( \frac{48 \pm 3\sqrt{481}}{25} \right)\)
\( \implies y = 4 - \frac{4(48 \pm 3\sqrt{481})}{75}\)
\( \implies y = \frac{300 - (192 \pm 12\sqrt{481})}{75}\)
\( \implies y = \frac{300 - 192 \mp 12\sqrt{481}}{75}\)
\( \implies y = \frac{108 \mp 12\sqrt{481}}{75}\)
\( \implies y = \frac{36 \mp 4\sqrt{481}}{25}\)
So, the two points of intersection are \( \left( \frac{48 + 3\sqrt{481}}{25}, \frac{36 - 4\sqrt{481}}{25} \right) \) and \( \left( \frac{48 - 3\sqrt{481}}{25}, \frac{36 + 4\sqrt{481}}{25} \right) \).
Let \( (x_1, y_1) = \left( \frac{48 + 3\sqrt{481}}{25}, \frac{36 - 4\sqrt{481}}{25} \right) \) and \( (x_2, y_2) = \left( \frac{48 - 3\sqrt{481}}{25}, \frac{36 + 4\sqrt{481}}{25} \right) \).
Now, we find the length of the intersecting chord using the distance formula \(D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\).
\(x_2 - x_1 = \frac{48 - 3\sqrt{481}}{25} - \frac{48 + 3\sqrt{481}}{25} = \frac{-6\sqrt{481}}{25}\)
\(y_2 - y_1 = \frac{36 + 4\sqrt{481}}{25} - \frac{36 - 4\sqrt{481}}{25} = \frac{8\sqrt{481}}{25}\)
Length of chord \(D = \sqrt{\left( \frac{-6\sqrt{481}}{25} \right)^2 + \left( \frac{8\sqrt{481}}{25} \right)^2}\)
\( \implies D = \sqrt{\frac{36 \cdot 481}{625} + \frac{64 \cdot 481}{625}}\)
\( \implies D = \sqrt{\frac{(36+64) \cdot 481}{625}}\)
\( \implies D = \sqrt{\frac{100 \cdot 481}{625}}\)
\( \implies D = \sqrt{\frac{4 \cdot 481}{25}}\)
\( \implies D = \frac{2\sqrt{481}}{5}\)
In simple words: We first find where the circle and the line meet by solving their equations together. This gives us two points. Then, we calculate the straight distance between these two points to find the length of the chord.

🎯 Exam Tip: Remember to use the quadratic formula carefully when solving for intersection points and double-check calculations for the distance formula.

Question 2. If circle \(x^2 + y^2 = a^2\) cuts an intercept of length \(2l\) at straight line \(mx + c\), then prove that \(c^2 = (1 + m^2)(a^2 – l^2)\).
Answer:
The equation of the circle is given as: \(x^2 + y^2 = a^2\).
From this, we know the center of the circle is \((0, 0)\) and its radius is \(a\).
The equation of the straight line is: \(y = mx + c\), which can be rewritten as \(mx - y + c = 0\).
The length of the intercept (chord) cut by the circle on the line is \(2l\). Therefore, half the length of the chord is \(l\).
The perpendicular distance \((d)\) from the center of the circle \((0,0)\) to the line \(mx - y + c = 0\) is given by the formula:
\(d = \frac{|A x_1 + B y_1 + C|}{\sqrt{A^2 + B^2}}\)
\( \implies d = \frac{|m(0) - 1(0) + c|}{\sqrt{m^2 + (-1)^2}}\)
\( \implies d = \frac{|c|}{\sqrt{m^2 + 1}}\)
Now, we can use the relationship between the radius \((a)\), the half-chord length \((l)\), and the perpendicular distance \((d)\) from the center to the chord. These form a right-angled triangle, so by the Pythagorean theorem:
\(a^2 = d^2 + l^2\)
Substitute the expression for \(d\):
\(a^2 = \left( \frac{|c|}{\sqrt{m^2 + 1}} \right)^2 + l^2\)
\( \implies a^2 = \frac{c^2}{m^2 + 1} + l^2\)
To prove the required equation, rearrange the terms:
\(a^2 - l^2 = \frac{c^2}{m^2 + 1}\)
\( \implies c^2 = (m^2 + 1)(a^2 - l^2)\)
Hence, the given condition is proved.
In simple words: We used the formula for the distance from the circle's center to the line and the Pythagorean theorem. The radius, the distance to the line, and half the chord make a right triangle. By putting these parts into the theorem, we showed the given relationship is true.

🎯 Exam Tip: Remember the formula for the perpendicular distance from a point to a line and how it relates to the radius and chord length in a circle.

Question 3. Find the length of intercept cut by circle \(x^2 + y^2 = c^2\) at line \( \frac{x}{a} + \frac{y}{b} = 1 \).
Answer:
The equation of the circle is given as: \(x^2 + y^2 = c^2\).
From this, the center of the circle is \((0, 0)\) and its radius \(R = c\).
The equation of the line is: \( \frac{x}{a} + \frac{y}{b} = 1 \).
To find the perpendicular distance from the center \((0,0)\) to the line, we first rewrite the line equation in the standard form \(Ax + By + C = 0\):
Multiply by \(ab\): \(bx + ay = ab\)
\( \implies bx + ay - ab = 0\)
The perpendicular distance \((d)\) from the center \((0,0)\) to the line \(bx + ay - ab = 0\) is:
\(d = \frac{|b(0) + a(0) - ab|}{\sqrt{b^2 + a^2}}\)
\( \implies d = \frac{|-ab|}{\sqrt{a^2 + b^2}}\)
\( \implies d = \frac{ab}{\sqrt{a^2 + b^2}}\)
Let the length of the intercept (chord) cut by the circle on the line be \(2L\). Then, half of the chord length is \(L\).
Using the Pythagorean theorem for the right-angled triangle formed by the radius \((R)\), the perpendicular distance \((d)\), and the half-chord length \((L)\):
\(R^2 = d^2 + L^2\)
Substitute \(R=c\) and the expression for \(d\):
\(c^2 = \left( \frac{ab}{\sqrt{a^2 + b^2}} \right)^2 + L^2\)
\( \implies c^2 = \frac{a^2 b^2}{a^2 + b^2} + L^2\)
Now, solve for \(L^2\):
\(L^2 = c^2 - \frac{a^2 b^2}{a^2 + b^2}\)
\( \implies L^2 = \frac{c^2(a^2 + b^2) - a^2 b^2}{a^2 + b^2}\)
The length of the intercept is \(2L\):
Length of intercept \( = 2\sqrt{c^2 - \frac{a^2 b^2}{a^2 + b^2}} \)
\( = 2\sqrt{\frac{c^2 a^2 + c^2 b^2 - a^2 b^2}{a^2 + b^2}}\)
In simple words: We found the distance from the center of the circle to the line. Then, using the Pythagorean theorem with the radius and this distance, we calculated half the length of the chord. Finally, we doubled it to get the full length of the intercept.

🎯 Exam Tip: When dealing with lines in intercept form, convert them to general form \(Ax+By+C=0\) to easily calculate the perpendicular distance from the origin.

Question 5. Find the condition, when
(i) Line \(y = mx + c\), touches the circle \((x – a)^2 + (y - b)^2 = r^2\)
(ii) Line \(lx + my + n = 0\) touches the circle \(x^2 + y^2 = a^2\)
Answer:
(i) For the line \(y = mx + c\) to touch the circle \((x – a)^2 + (y - b)^2 = r^2\):
The equation of the circle is \((x – a)^2 + (y - b)^2 = r^2\). Its center is \((a, b)\) and its radius is \(r\).
The equation of the line is \(y = mx + c\), which can be written as \(mx - y + c = 0\).
For the line to be tangent to the circle, the perpendicular distance from the center of the circle to the line must be equal to the radius of the circle.
The perpendicular distance \((d)\) from \((a, b)\) to \(mx - y + c = 0\) is:
\(d = \frac{|m(a) - b + c|}{\sqrt{m^2 + (-1)^2}}\)
\( \implies d = \frac{|ma - b + c|}{\sqrt{m^2 + 1}}\)
Since \(d = r\):
\(r = \frac{|ma - b + c|}{\sqrt{m^2 + 1}}\)
Square both sides to get the condition:
\(r^2 = \frac{(ma - b + c)^2}{m^2 + 1}\)
\( \implies (1 + m^2)r^2 = (ma - b + c)^2\)
This is the required condition for tangency. The source then shows an expansion of this condition:
\( (1 + m^2)r^2 = m^2 a^2 + b^2 + c^2 - 2mab - 2bc + 2mac \)
Rearranging terms (as shown in the source):
\( m^2(a^2 - r^2) + 2a(c - b)m + (c - b)^2 = r^2 \)

(ii) For the line \(lx + my + n = 0\) to touch the circle \(x^2 + y^2 = a^2\):
The equation of the circle is \(x^2 + y^2 = a^2\). Its center is \((0, 0)\) and its radius is \(a\).
The equation of the line is \(lx + my + n = 0\).
For tangency, the perpendicular distance from the center \((0, 0)\) to the line must equal the radius \(a\).
The perpendicular distance \((d)\) from \((0, 0)\) to \(lx + my + n = 0\) is:
\(d = \frac{|l(0) + m(0) + n|}{\sqrt{l^2 + m^2}}\)
\( \implies d = \frac{|n|}{\sqrt{l^2 + m^2}}\)
Since \(d = a\):
\(a = \frac{|n|}{\sqrt{l^2 + m^2}}\)
Square both sides to get the condition:
\(a^2 = \frac{n^2}{l^2 + m^2}\)
\( \implies a^2(l^2 + m^2) = n^2\)
This is the required condition.
In simple words: A line touches a circle if the distance from the circle's center to the line is exactly the same as the circle's radius. We use the distance formula and set it equal to the radius to find the condition.

🎯 Exam Tip: Always remember that the condition for a line to be tangent to a circle is that the perpendicular distance from the center of the circle to the line must be equal to the radius of the circle.

Question 6. (i) Find the equation of tangent of the circle \(x^2 + y^2 = 64\) which passes through point \((4, 7)\).
(ii) Find the equation of tangent of the circle \(x^2 + y^2 – 4\) which makes an angle of \(60^\circ\) with x-axis.
Answer:
(i) To find the equation of the tangent to the circle \(x^2 + y^2 = 64\) passing through point \((4, 7)\):
The given circle equation is \(x^2 + y^2 = 64\), which means its center is \((0, 0)\) and its radius is \(r = \sqrt{64} = 8\).
Let the equation of the tangent line be \(y - y_1 = m(x - x_1)\). Since it passes through \((4, 7)\):
\(y - 7 = m(x - 4)\)
\( \implies y - 7 = mx - 4m\)
\( \implies mx - y + (7 - 4m) = 0\)
For this line to be tangent to the circle, the perpendicular distance from the center \((0, 0)\) to the line must be equal to the radius \(8\).
\(d = \frac{|A x_1 + B y_1 + C|}{\sqrt{A^2 + B^2}}\)
\( \implies 8 = \frac{|m(0) - 1(0) + (7 - 4m)|}{\sqrt{m^2 + (-1)^2}}\)
\( \implies 8 = \frac{|7 - 4m|}{\sqrt{m^2 + 1}}\)
Square both sides to remove the square root and absolute value:
\(64 = \frac{(7 - 4m)^2}{m^2 + 1}\)
\( \implies 64(m^2 + 1) = (7 - 4m)^2\)
\( \implies 64m^2 + 64 = 49 - 56m + 16m^2\)
Rearrange the terms to form a quadratic equation:
\(64m^2 - 16m^2 + 56m + 64 - 49 = 0\)
\( \implies 48m^2 + 56m + 15 = 0\)
Factor this quadratic equation:
\(48m^2 + 36m + 20m + 15 = 0\)
\( \implies 12m(4m + 3) + 5(4m + 3) = 0\)
\( \implies (12m + 5)(4m + 3) = 0\)
This gives two possible values for the slope \(m\):
If \(12m + 5 = 0 \implies m = -\frac{5}{12}\)
If \(4m + 3 = 0 \implies m = -\frac{3}{4}\)
Now, substitute each \(m\) value back into the tangent equation \(y - 7 = m(x - 4)\):
For \(m = -\frac{5}{12}\):
\(y - 7 = -\frac{5}{12}(x - 4)\)
\( \implies 12(y - 7) = -5(x - 4)\)
\( \implies 12y - 84 = -5x + 20\)
\( \implies 5x + 12y - 104 = 0\)
For \(m = -\frac{3}{4}\):
\(y - 7 = -\frac{3}{4}(x - 4)\)
\( \implies 4(y - 7) = -3(x - 4)\)
\( \implies 4y - 28 = -3x + 12\)
\( \implies 3x + 4y - 40 = 0\)
These are the two equations of the tangents.

(ii) To find the equation of the tangent to the circle \(x^2 + y^2 = 4\) which makes an angle of \(60^\circ\) with the x-axis:
The given circle equation is \(x^2 + y^2 = 4\), which can be written as \(x^2 + y^2 = 2^2\). Its center is \((0, 0)\) and its radius is \(a = 2\).
The tangent line makes an angle of \(60^\circ\) with the x-axis, so its slope is \(m = \tan(60^\circ) = \sqrt{3}\).
The general equation of a tangent to a circle \(x^2 + y^2 = a^2\) with slope \(m\) is \(y = mx \pm a\sqrt{1 + m^2}\).
Substitute the values of \(a\) and \(m\):
\(y = \sqrt{3}x \pm 2\sqrt{1 + (\sqrt{3})^2}\)
\( \implies y = \sqrt{3}x \pm 2\sqrt{1 + 3}\)
\( \implies y = \sqrt{3}x \pm 2\sqrt{4}\)
\( \implies y = \sqrt{3}x \pm 2(2)\)
\( \implies y = \sqrt{3}x \pm 4\)
These are the two required equations of the tangents.
In simple words: For part (i), we used the fact that the distance from the center of the circle to the tangent line equals the radius. This helped us find the slopes of the tangent lines. For part (ii), we used a standard formula for the tangent line when we know its slope and the circle's radius.

🎯 Exam Tip: Remember the condition for tangency (distance from center = radius) and the standard tangent equation \(y = mx \pm a\sqrt{1 + m^2}\) for circles centered at the origin.

Question 7. Find the value of c, where line \(y = c\), touches the circle \(x^2 + y^2 - 2x + 2y - 2 = 0\) at point \((1, 1)\).
Answer:
The given equation of the circle is \(x^2 + y^2 - 2x + 2y - 2 = 0\).
We compare this with the general equation of a circle \(x^2 + y^2 + 2gx + 2fy + k = 0\).
From comparison:
\(2g = -2 \implies g = -1\)
\(2f = 2 \implies f = 1\)
\(k = -2\)
The center of the circle is \((-g, -f) = (1, -1)\).
The radius of the circle is \(R = \sqrt{g^2 + f^2 - k} = \sqrt{(-1)^2 + (1)^2 - (-2)}\)
\( \implies R = \sqrt{1 + 1 + 2}\)
\( \implies R = \sqrt{4}\)
\( \implies R = 2\)
The line is given as \(y = c\). This line touches the circle at the point \((1, 1)\).
Since the point of tangency \((1, 1)\) lies on the line \(y = c\), it must satisfy the equation of the line.
Substitute \(x=1, y=1\) into \(y=c\):
\(1 = c\)
Thus, the value of \(c\) is \(1\).
Alternatively, we can find the equation of the tangent at \((1, 1)\) to the circle \(x^2 + y^2 - 2x + 2y - 2 = 0\).
The equation of the tangent at \((x_1, y_1)\) to a general circle is \(x x_1 + y y_1 + g(x + x_1) + f(y + y_1) + k = 0\).
Here, \((x_1, y_1) = (1, 1)\), \(g = -1\), \(f = 1\), \(k = -2\).
\(x(1) + y(1) + (-1)(x + 1) + (1)(y + 1) + (-2) = 0\)
\( \implies x + y - x - 1 + y + 1 - 2 = 0\)
\( \implies 2y - 2 = 0\)
\( \implies 2y = 2\)
\( \implies y = 1\)
Comparing this tangent equation \(y=1\) with the given line equation \(y=c\), we find \(c=1\).
In simple words: Since the line touches the circle at a specific point, that point must be on the line. By putting the coordinates of the touching point into the line's equation, we can find the value of 'c'.

🎯 Exam Tip: If a line touches a circle at a specific point, that point must lie on both the circle and the line. Also, know how to find the center and radius from the general circle equation and the tangent equation at a point.

Question 8. Find the equation of tangent at point \((5, 12)\) and \((12, – 5)\) at circle \(x^2 + y^2 = 169\). Prove that they will be perpendicular to each other. Also find the coordinates of intersection point.
Answer:
The given equation of the circle is \(x^2 + y^2 = 169\). Its center is \((0, 0)\) and radius is \(\sqrt{169} = 13\).
The equation of the tangent at a point \((x_1, y_1)\) to the circle \(x^2 + y^2 = a^2\) is given by \(x x_1 + y y_1 = a^2\).

**Tangent at point \((5, 12)\):**
Here, \((x_1, y_1) = (5, 12)\) and \(a^2 = 169\).
The equation of the tangent is: \(x(5) + y(12) = 169\)
\( \implies 5x + 12y - 169 = 0\) ...(ii)
The slope of this tangent \(m_1 = -\frac{\text{Coefficient of } x}{\text{Coefficient of } y} = -\frac{5}{12}\).

**Tangent at point \((12, -5)\):**
Here, \((x_1, y_1) = (12, -5)\) and \(a^2 = 169\).
The equation of the tangent is: \(x(12) + y(-5) = 169\)
\( \implies 12x - 5y - 169 = 0\) ...(iii)
The slope of this tangent \(m_2 = -\frac{\text{Coefficient of } x}{\text{Coefficient of } y} = -\frac{12}{-5} = \frac{12}{5}\).

**Proof of Perpendicularity:**
To check if the two tangents are perpendicular, we multiply their slopes:
\(m_1 \cdot m_2 = \left(-\frac{5}{12}\right) \cdot \left(\frac{12}{5}\right) = -1\)
Since the product of their slopes is \(-1\), the two tangent lines are perpendicular to each other.

**Coordinates of Intersection Point:**
To find the intersection point, we solve equations (ii) and (iii) simultaneously:
1. \(5x + 12y = 169\) (from (ii))
2. \(12x - 5y = 169\) (from (iii))
Multiply equation (ii) by 5 and equation (iii) by 12 to eliminate \(y\):
\(5 \times (5x + 12y = 169) \implies 25x + 60y = 845\) ...(iv)
\(12 \times (12x - 5y = 169) \implies 144x - 60y = 2028\) ...(v)
Add equation (iv) and (v):
\((25x + 144x) + (60y - 60y) = 845 + 2028\)
\( \implies 169x = 2873\)
\( \implies x = \frac{2873}{169}\)
\( \implies x = 17\)
Substitute the value of \(x=17\) into equation (ii):
\(5(17) + 12y = 169\)
\( \implies 85 + 12y = 169\)
\( \implies 12y = 169 - 85\)
\( \implies 12y = 84\)
\( \implies y = \frac{84}{12}\)
\( \implies y = 7\)
So, the coordinates of the intersection point are \((17, 7)\).
In simple words: We found the equations for two lines that touch the circle at specific points. We showed they are at right angles to each other by checking their slopes. Then, we solved their equations to find where they cross paths.

🎯 Exam Tip: Remember the formula for the tangent to a circle at a given point, the condition for perpendicular lines (\(m_1m_2 = -1\)), and how to solve simultaneous linear equations to find intersection points.