OP Malhotra Class 11 Maths Solutions Chapter 17 Circle Exercise 17 (D)

Question 1. Find the equations of the tangent to the circle \( 2x^2 + 2y^2 = 5 \) which are perpendicular to \( y = 2x \).
Answer: First, we write the given equation of the circle in its standard form. The equation of the circle is \( 2x^2 + 2y^2 = 5 \). Divide by 2 to get: \( x^2 + y^2 = \frac{5}{2} \) ...(1) This circle has its center at \( (0, 0) \) and its radius \( r = \sqrt{\frac{5}{2}} \). Next, we look at the given line \( y = 2x \), which can be written as \( 2x - y = 0 \) ...(2). We need to find the equation of a line perpendicular to this line. The slope of line (2) is 2. The slope of a line perpendicular to it will be \( -\frac{1}{2} \). So, the equation of a line perpendicular to line (2) is of the form \( y = -\frac{1}{2}x + k' \) or \( x + 2y + k = 0 \) ...(3). Now, if line (3) is a tangent to circle (1), the distance from the center of the circle \( (0, 0) \) to line (3) must be equal to the radius of the circle. Using the distance formula from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \), which is \( \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \):
\( \implies \) \( \frac{|1(0) + 2(0) + k|}{\sqrt{1^2 + 2^2}} = \sqrt{\frac{5}{2}} \)
\( \implies \) \( \frac{|k|}{\sqrt{1 + 4}} = \sqrt{\frac{5}{2}} \)
\( \implies \) \( \frac{|k|}{\sqrt{5}} = \frac{\sqrt{5}}{\sqrt{2}} \)
\( \implies \) \( |k| = \frac{\sqrt{5} \times \sqrt{5}}{\sqrt{2}} \)
\( \implies \) \( |k| = \frac{5}{\sqrt{2}} \)
\( \implies \) \( k = \pm \frac{5}{\sqrt{2}} \) Finally, we substitute these values of \( k \) back into the equation of the line \( x + 2y + k = 0 \). So, the required equations of the tangents are \( x + 2y \pm \frac{5}{\sqrt{2}} = 0 \). These two equations represent the two possible tangent lines.In simple words: First, we change the circle's equation to a simple form. Then, we find the line that is exactly perpendicular to the given line. For this new line to be a tangent to the circle, its distance from the center of the circle must be the same as the circle's radius. We use this rule to find the missing number in the line's equation, which gives us two possible tangent lines.

🎯 Exam Tip: Remember to always convert the circle's equation to the standard form \( x^2 + y^2 = r^2 \) or \( (x-h)^2 + (y-k)^2 = r^2 \) to easily identify the center and radius. The condition that the perpendicular distance from the center to the tangent equals the radius is crucial.

Question 2. Find the equations of the tangents to the circle \( x^2 + y^2 – 8y – 8 = 0 \) which are parallel to the line \( 5x – 2y = 2 \).
Answer: First, let's find the center and radius of the given circle. The equation of the circle is \( x^2 + y^2 – 8y – 8 = 0 \). We can rewrite this by completing the square for the y terms: \( x^2 + (y^2 – 8y + 16) – 8 – 16 = 0 \) \( x^2 + (y - 4)^2 = 24 \) ...(1) So, the center of this circle is \( C(0, 4) \) and its radius is \( r = \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6} \). Next, we need the equations of lines parallel to the given line \( 5x – 2y = 2 \). A line parallel to \( 5x – 2y - 2 = 0 \) will have the same slope, so its equation will be of the form \( 5x – 2y + k = 0 \) ...(2). For line (2) to be a tangent to circle (1), the perpendicular distance from the center of the circle \( C(0, 4) \) to line (2) must be equal to the radius \( r = \sqrt{24} \). Using the distance formula:
\( \implies \) \( \frac{|5(0) - 2(4) + k|}{\sqrt{5^2 + (-2)^2}} = \sqrt{24} \)
\( \implies \) \( \frac{|-8 + k|}{\sqrt{25 + 4}} = \sqrt{24} \)
\( \implies \) \( \frac{|k - 8|}{\sqrt{29}} = \sqrt{24} \)
\( \implies \) \( |k - 8| = \sqrt{24}\sqrt{29} \)
\( \implies \) \( k - 8 = \pm \sqrt{24 \times 29} \)
\( \implies \) \( k - 8 = \pm \sqrt{696} \)
\( \implies \) \( k = 8 \pm \sqrt{696} \) We can simplify \( \sqrt{696} = \sqrt{4 \times 174} = 2\sqrt{174} \). So, \( k = 8 \pm 2\sqrt{174} \). Substitute these values of \( k \) back into the equation of the line \( 5x – 2y + k = 0 \). The required equations of the tangents are \( 5x – 2y + (8 \pm 2\sqrt{174}) = 0 \).In simple words: First, we find the center and radius of the circle. Then, we find lines that have the same slope as the given line (because they are parallel). For these parallel lines to be tangents, their distance from the circle's center must be equal to the circle's radius. We use this condition to find the unknown part of the line's equation, giving us the two tangent lines.

🎯 Exam Tip: When finding equations of parallel lines, only the constant term changes. For perpendicular lines, the coefficients of x and y swap, and one sign changes, along with a new constant term. Always complete the square to find the center and radius of a circle if it's not in standard form.

Question 3. Find the equation of the circle which has extremities of a diameter the origin and the point \( (2, – 4) \). Find also the equations of the tangents to the circle which are parallel to this diameter.
Answer: Let the origin be \( A(0, 0) \) and the other point be \( B(2, -4) \). These are the endpoints of a diameter. The equation of a circle with diameter endpoints \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by \( (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \). So, the equation of the circle is: \( (x - 0)(x - 2) + (y - 0)(y - (-4)) = 0 \) \( x(x - 2) + y(y + 4) = 0 \) \( x^2 - 2x + y^2 + 4y = 0 \) Rearranging, we get: \( x^2 + y^2 - 2x + 4y = 0 \) ...(1). To find the center and radius of this circle: \( (x^2 - 2x + 1) + (y^2 + 4y + 4) = 1 + 4 \) \( (x - 1)^2 + (y + 2)^2 = 5 \) The center of the circle is \( C(1, -2) \) and the radius is \( r = \sqrt{5} \). Next, we find the equation of the diameter that passes through \( (0, 0) \) and \( (2, -4) \). Using the two-point form \( y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \): \( y - 0 = \frac{-4 - 0}{2 - 0}(x - 0) \) \( y = \frac{-4}{2}x \) \( y = -2x \) So, the equation of the diameter is \( 2x + y = 0 \) ...(2). Now, we need to find the equations of tangents parallel to this diameter. A line parallel to \( 2x + y = 0 \) will have the form \( 2x + y + k = 0 \) ...(3). For line (3) to be a tangent to the circle, the perpendicular distance from the center of the circle \( C(1, -2) \) to line (3) must be equal to the radius \( r = \sqrt{5} \). Using the distance formula:
\( \implies \) \( \frac{|2(1) + 1(-2) + k|}{\sqrt{2^2 + 1^2}} = \sqrt{5} \)
\( \implies \) \( \frac{|2 - 2 + k|}{\sqrt{4 + 1}} = \sqrt{5} \)
\( \implies \) \( \frac{|k|}{\sqrt{5}} = \sqrt{5} \)
\( \implies \) \( |k| = \sqrt{5} \times \sqrt{5} \)
\( \implies \) \( |k| = 5 \)
\( \implies \) \( k = \pm 5 \) Substitute these values of \( k \) back into the equation of the line \( 2x + y + k = 0 \). The required equations of the tangents are \( 2x + y \pm 5 = 0 \).In simple words: First, we find the equation of the circle using the given diameter endpoints. Then, we find the center and radius of this circle. Next, we determine the equation of the diameter itself. We then look for lines that are parallel to this diameter. For these lines to be tangents to the circle, their distance from the circle's center must be equal to the circle's radius. This helps us find the specific equations for the tangent lines.

🎯 Exam Tip: The formula \( (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \) is very useful for finding the circle's equation when diameter endpoints are given. Remember that a tangent is a line whose perpendicular distance from the center equals the radius.

Question 4. Show that, whatever be the value of a, the lines \( x \cos \alpha + y \sin \alpha = a \) and \( x \sin \alpha – y \cos \alpha = a \) are tangents to the circle \( x^2 + y^2 = a^2 \). Hence obtain the locus of the points from which perpendicular tangents can be drawn to the circle \( x^2 + y^2 = a^2 \).
Answer: Let's analyze the given circle and lines. The equation of the circle is \( x^2 + y^2 = a^2 \) ...(1). The center of this circle is \( (0, 0) \) and its radius is \( r = \sqrt{a^2} = |a| \). For simplicity, we can assume \( a > 0 \), so \( r = a \). The given lines are: Line (2): \( x \cos \alpha + y \sin \alpha - a = 0 \) Line (3): \( x \sin \alpha – y \cos \alpha - a = 0 \) To show that these lines are tangents, we need to prove that the perpendicular distance from the center \( (0, 0) \) to each line is equal to the radius \( a \). For Line (2): Distance \( d_1 = \frac{|(0)\cos \alpha + (0)\sin \alpha - a|}{\sqrt{(\cos \alpha)^2 + (\sin \alpha)^2}} \)
\( \implies \) \( d_1 = \frac{|-a|}{\sqrt{\cos^2 \alpha + \sin^2 \alpha}} \) Since \( \cos^2 \alpha + \sin^2 \alpha = 1 \):
\( \implies \) \( d_1 = \frac{|-a|}{\sqrt{1}} = |a| = a \) (assuming \( a > 0 \)). For Line (3): Distance \( d_2 = \frac{|(0)\sin \alpha - (0)\cos \alpha - a|}{\sqrt{(\sin \alpha)^2 + (-\cos \alpha)^2}} \)
\( \implies \) \( d_2 = \frac{|-a|}{\sqrt{\sin^2 \alpha + \cos^2 \alpha}} \)
\( \implies \) \( d_2 = \frac{|-a|}{\sqrt{1}} = |a| = a \) (assuming \( a > 0 \)). Since \( d_1 = d_2 = a \), which is the radius of the circle, both lines (2) and (3) are indeed tangents to the given circle, regardless of the value of \( \alpha \). Now, let's find the locus of the points from which perpendicular tangents can be drawn. This is known as the Director Circle. First, we find the slopes of the lines: For Line (2): \( m_1 = -\frac{\text{coefficient of x}}{\text{coefficient of y}} = -\frac{\cos \alpha}{\sin \alpha} = -\cot \alpha \) For Line (3): \( m_2 = -\frac{\text{coefficient of x}}{\text{coefficient of y}} = -\frac{\sin \alpha}{-\cos \alpha} = \frac{\sin \alpha}{\cos \alpha} = \tan \alpha \) Now, let's check if these tangents are perpendicular: \( m_1 m_2 = (-\cot \alpha)(\tan \alpha) = (- \frac{\cos \alpha}{\sin \alpha}) (\frac{\sin \alpha}{\cos \alpha}) = -1 \) Since \( m_1 m_2 = -1 \), the two tangents are perpendicular to each other. To find the locus of the point of intersection of these perpendicular tangents, we can eliminate \( \alpha \) from the equations of the lines. Let \( (x, y) \) be the point of intersection. Then \( x \cos \alpha + y \sin \alpha = a \) ...(A) and \( x \sin \alpha – y \cos \alpha = a \) ...(B). We want to eliminate \( \alpha \). Square both equations and add them: \( (x \cos \alpha + y \sin \alpha)^2 = a^2 \) \( x^2 \cos^2 \alpha + y^2 \sin^2 \alpha + 2xy \cos \alpha \sin \alpha = a^2 \) \( (x \sin \alpha – y \cos \alpha)^2 = a^2 \) \( x^2 \sin^2 \alpha + y^2 \cos^2 \alpha - 2xy \sin \alpha \cos \alpha = a^2 \) Adding these two squared equations: \( (x^2 \cos^2 \alpha + y^2 \sin^2 \alpha + 2xy \cos \alpha \sin \alpha) + (x^2 \sin^2 \alpha + y^2 \cos^2 \alpha - 2xy \sin \alpha \cos \alpha) = a^2 + a^2 \) \( x^2 (\cos^2 \alpha + \sin^2 \alpha) + y^2 (\sin^2 \alpha + \cos^2 \alpha) = 2a^2 \) Since \( \cos^2 \alpha + \sin^2 \alpha = 1 \): \( x^2 (1) + y^2 (1) = 2a^2 \) \( x^2 + y^2 = 2a^2 \) This equation, \( x^2 + y^2 = 2a^2 \), represents the locus of the points from which perpendicular tangents can be drawn to the circle \( x^2 + y^2 = a^2 \). This is known as the Director Circle. The radius of the director circle is \( \sqrt{2}a \).In simple words: First, we show that the distance from the center of the circle to each line is equal to the circle's radius. This proves both lines are tangents. Next, we find the slopes of these lines and show that they are perpendicular to each other. Finally, we find the path (locus) of all points from which two such perpendicular tangent lines can be drawn. This path turns out to be another circle, called the Director Circle, whose equation is \( x^2 + y^2 = 2a^2 \).

🎯 Exam Tip: To show a line is tangent to a circle, prove that the perpendicular distance from the circle's center to the line equals the circle's radius. The locus of points from which perpendicular tangents can be drawn to a circle \( x^2+y^2=r^2 \) is always the director circle \( x^2+y^2=2r^2 \).

Question 5. Find the locus of the feet of the perpendiculars drawn from the point \( (b, 0) \) on tangents to the circle \( x^2 + y^2 = a^2 \).
Answer: Let the equation of the circle be \( x^2 + y^2 = a^2 \). The center of this circle is \( (0, 0) \) and its radius is \( a \). The equation of any tangent to the circle \( x^2 + y^2 = a^2 \) is given by \( y = mx \pm a\sqrt{1 + m^2} \) ...(1). Let \( P \) be the point \( (b, 0) \). Let \( Q(x, y) \) be the foot of the perpendicular drawn from \( P \) to the tangent line (1). The line \( PQ \) is perpendicular to the tangent. The slope of the tangent (1) is \( m \). So, the slope of the line \( PQ \) must be \( -\frac{1}{m} \). The equation of the line \( PQ \) passing through \( P(b, 0) \) with slope \( -\frac{1}{m} \) is: \( y - 0 = -\frac{1}{m}(x - b) \)
\( \implies \) \( my = -x + b \)
\( \implies \) \( x + my = b \) ...(2). Now, to find the locus of the feet of the perpendiculars (point \( Q(x,y) \)), we need to eliminate \( m \) from equations (1) and (2). From equation (2), we can write \( my = b - x \). If \( y \neq 0 \), then \( m = \frac{b - x}{y} \). Substitute this value of \( m \) into equation (1): \( y = \left(\frac{b - x}{y}\right)x \pm a\sqrt{1 + \left(\frac{b - x}{y}\right)^2} \) Multiply by \( y \) (assuming \( y \neq 0 \)): \( y^2 = (b - x)x \pm ay\sqrt{1 + \frac{(b - x)^2}{y^2}} \) \( y^2 = (b - x)x \pm a\sqrt{y^2 + (b - x)^2} \) Rearrange to isolate the square root term: \( y^2 - (b - x)x = \pm a\sqrt{y^2 + (b - x)^2} \) Square both sides to remove the square root and the \( \pm \) sign: \( (y^2 - bx + x^2)^2 = a^2 (y^2 + (b - x)^2) \) \( (x^2 + y^2 - bx)^2 = a^2 (y^2 + b^2 - 2bx + x^2) \) \( (x^2 + y^2 - bx)^2 = a^2 (x^2 + y^2 - 2bx + b^2) \) This is the required equation of the locus. It represents the pedal curve of the circle with respect to the point \( (b, 0) \).In simple words: We start with the general equation of a tangent line to the circle. Then, we write the equation of the line that goes from the point \( (b, 0) \) and is perpendicular to the tangent line. The point where these two lines meet is called the 'foot of the perpendicular'. To find the path (locus) of all such feet, we remove the slope 'm' from both equations. The resulting equation, \( (x^2 + y^2 - bx)^2 = a^2 (x^2 + y^2 - 2bx + b^2) \), shows the shape of this path.

🎯 Exam Tip: The key to locus problems involving perpendiculars to tangents is to use the general form of a tangent, the condition for perpendicular lines, and then eliminate the parameter (like 'm' here) to get the equation of the locus. Squaring both sides is often needed to get rid of square roots.

Question 6. (i) Find the equation of that chord of the circle \( x^2 + y^2 = 15 \), which is bisected at the point \( (3, 2) \). (ii) Find the locus of mid-points of all chords of the circle \( x^2 + y^2 = 15 \) that pass through the point \( (3, 4) \).
Answer:
(i) Let the equation of the circle be \( x^2 + y^2 = 15 \) ...(1). The center of this circle is \( C(0, 0) \) and its radius is \( \sqrt{15} \). Let \( M(3, 2) \) be the midpoint of the chord \( AB \). The line segment \( CM \) connects the center of the circle to the midpoint of the chord. This line \( CM \) is always perpendicular to the chord \( AB \). Slope of \( CM = \frac{2 - 0}{3 - 0} = \frac{2}{3} \). Since chord \( AB \) is perpendicular to \( CM \), the slope of chord \( AB \) is \( -\frac{1}{\text{slope of CM}} = -\frac{1}{(2/3)} = -\frac{3}{2} \). Now we have the midpoint \( M(3, 2) \) and the slope \( -\frac{3}{2} \) for the chord \( AB \). Using the point-slope form \( y - y_1 = m(x - x_1) \) for the equation of the chord: \( y - 2 = -\frac{3}{2}(x - 3) \) Multiply by 2: \( 2(y - 2) = -3(x - 3) \) \( 2y - 4 = -3x + 9 \) Rearrange the terms: \( 3x + 2y - 4 - 9 = 0 \) \( 3x + 2y - 13 = 0 \) This is the required equation of the chord.
(ii) The equation of the circle is \( x^2 + y^2 = 15 \). The center is \( C(0, 0) \). Let \( M(\alpha, \beta) \) be the midpoint of any chord \( QR \) of the circle. The line joining the center \( C \) to the midpoint \( M \) of the chord is perpendicular to the chord \( QR \). Slope of \( CM = \frac{\beta - 0}{\alpha - 0} = \frac{\beta}{\alpha} \). Since \( QR \perp CM \), the slope of chord \( QR = -\frac{1}{\text{slope of CM}} = -\frac{\alpha}{\beta} \). The equation of the chord \( QR \) passing through \( M(\alpha, \beta) \) with slope \( -\frac{\alpha}{\beta} \) is: \( y - \beta = -\frac{\alpha}{\beta}(x - \alpha) \) Multiply by \( \beta \): \( \beta(y - \beta) = -\alpha(x - \alpha) \) \( \beta y - \beta^2 = -\alpha x + \alpha^2 \) Rearrange: \( \alpha x + \beta y = \alpha^2 + \beta^2 \) ...(A). We are given that this chord \( QR \) passes through the point \( P(3, 4) \). So, the point \( (3, 4) \) must satisfy equation (A): \( \alpha(3) + \beta(4) = \alpha^2 + \beta^2 \) \( 3\alpha + 4\beta = \alpha^2 + \beta^2 \) Rearranging the terms: \( \alpha^2 + \beta^2 - 3\alpha - 4\beta = 0 \) To find the locus of the midpoint \( M(\alpha, \beta) \), we replace \( \alpha \) with \( x \) and \( \beta \) with \( y \). So, the locus of the midpoints is \( x^2 + y^2 - 3x - 4y = 0 \). This equation represents another circle passing through the origin.In simple words: (i) We find the center of the circle. Since the line from the center to the midpoint of a chord is always perpendicular to the chord, we can find the slope of the chord. With the midpoint and its slope, we write the chord's equation. (ii) For any chord, if its midpoint is \( (\alpha, \beta) \), the line from the circle's center to this midpoint is perpendicular to the chord. We write the equation of such a chord. Since all these chords pass through a specific point \( (3, 4) \), we use this information to find the relationship between \( \alpha \) and \( \beta \), which gives us the path (locus) of all such midpoints.

🎯 Exam Tip: For problems involving chords and midpoints, remember the geometric property that the line segment from the center of the circle to the midpoint of a chord is perpendicular to the chord. This relationship is key to finding slopes and equations. For locus problems, substitute the general coordinates of the point \( (\alpha, \beta) \) into the derived relation and then replace \( \alpha \) with \( x \) and \( \beta \) with \( y \).

Question 7. Find the locus of the middle points of the chords of the circle \( x^2 + y^2 = 4(y + 1) \) drawn through the origin.
Answer: First, let's rewrite the equation of the circle to find its center and radius. The given equation is \( x^2 + y^2 = 4(y + 1) \). \( x^2 + y^2 - 4y - 4 = 0 \) To find the center, we complete the square for the y terms: \( x^2 + (y^2 - 4y + 4) - 4 - 4 = 0 \) \( x^2 + (y - 2)^2 = 8 \) ...(1) The center of this circle is \( C(0, 2) \) and its radius is \( \sqrt{8} = 2\sqrt{2} \). Let \( M(\alpha, \beta) \) be the midpoint of any chord \( QR \) of the circle. We are given that these chords pass through the origin \( O(0, 0) \). So, the chord \( QR \) passes through \( O(0, 0) \) and its midpoint is \( M(\alpha, \beta) \). The line segment \( OM \) is part of the chord \( QR \). We know that the line segment from the center of the circle \( C(0, 2) \) to the midpoint of the chord \( M(\alpha, \beta) \) is perpendicular to the chord \( QR \). So, \( CM \perp OM \). This means the product of their slopes must be -1. Slope of \( CM = \frac{\beta - 2}{\alpha - 0} = \frac{\beta - 2}{\alpha} \). Slope of \( OM = \frac{\beta - 0}{\alpha - 0} = \frac{\beta}{\alpha} \). Since \( CM \perp OM \): \( (\text{Slope of CM}) \times (\text{Slope of OM}) = -1 \)
\( \implies \) \( \left(\frac{\beta - 2}{\alpha}\right) \times \left(\frac{\beta}{\alpha}\right) = -1 \)
\( \implies \) \( \frac{\beta(\beta - 2)}{\alpha^2} = -1 \)
\( \implies \) \( \beta^2 - 2\beta = -\alpha^2 \)
\( \implies \) \( \alpha^2 + \beta^2 - 2\beta = 0 \) To find the locus of the midpoint \( M(\alpha, \beta) \), we replace \( \alpha \) with \( x \) and \( \beta \) with \( y \). So, the locus of the middle points of the chords is \( x^2 + y^2 - 2y = 0 \). This equation represents another circle passing through the origin.In simple words: First, we find the center of the main circle. Then, we consider a chord whose midpoint is \( (\alpha, \beta) \) and that passes through the origin \( (0, 0) \). We know that the line from the circle's center to this midpoint is perpendicular to the chord itself. Using the slopes of these two perpendicular lines, we set their product to -1. This helps us find a relationship between \( \alpha \) and \( \beta \). Changing \( \alpha \) to \( x \) and \( \beta \) to \( y \) gives us the equation of the path (locus) of all such midpoints.

🎯 Exam Tip: When a chord passes through a fixed point, and you need the locus of its midpoint, remember that the line from the circle's center to the midpoint of the chord is perpendicular to the chord. The fixed point will be on the chord, making it possible to relate the midpoint's coordinates using slopes.