Samacheer Kalvi Class 11 Business Maths Solutions Chapter 3 Analytical Geometry Exercise 3.5

Question 1. Find the equation of the tangent to the circle \( x^2 + y^2 - 4x + 4y - 8 = 0 \) at (-2, -2).
Answer: The equation of the circle is given as \( x^2 + y^2 - 4x + 4y - 8 = 0 \). The point where the tangent touches is \( (x_1, y_1) = (-2, -2) \). The general equation of a tangent to a circle at a point \( (x_1, y_1) \) is \( xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0 \). From the given circle equation, \( 2g = -4 \implies g = -2 \) and \( 2f = 4 \implies f = 2 \), and \( c = -8 \).
So, the equation of the tangent is:
\( x(-2) + y(-2) - 2(x + (-2)) + 2(y + (-2)) - 8 = 0 \)
\( \implies -2x - 2y - 2(x - 2) + 2(y - 2) - 8 = 0 \)
\( \implies -2x - 2y - 2x + 4 + 2y - 4 - 8 = 0 \)
\( \implies -4x - 8 = 0 \)
\( \implies 4x + 8 = 0 \)
\( \implies x + 2 = 0 \)
The tangent line only touches the circle at one specific point. This line helps to understand the instantaneous direction of the curve at that point.
In simple words: We use a special formula to find the line that just touches the circle at the given point (-2, -2). The line we found is \( x + 2 = 0 \).

🎯 Exam Tip: Remember the formula for the equation of a tangent to a circle at a given point \( (x_1, y_1) \): \( xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0 \). Make sure to correctly identify the values of \( g, f, \) and \( c \) from the circle's equation.

Question 2. Determine whether the points P(1, 0), Q(2, 1) and R(2, 3) lie outside the circle, on the circle or inside the circle \( x^2 + y^2 - 4x - 6y + 9 = 0 \).
Answer: The equation of the circle is \( x^2 + y^2 - 4x - 6y + 9 = 0 \). To determine if a point \( (x_1, y_1) \) is outside, on, or inside the circle, we substitute the coordinates into the expression \( S_1 = x_1^2 + y_1^2 - 4x_1 - 6y_1 + 9 \).
If \( S_1 > 0 \), the point lies outside the circle.
If \( S_1 = 0 \), the point lies on the circle.
If \( S_1 < 0 \), the point lies inside the circle.

For point P(1, 0):
\( S_1 = 1^2 + 0^2 - 4(1) - 6(0) + 9 = 1 + 0 - 4 - 0 + 9 = 6 \)
Since \( S_1 = 6 > 0 \), point P lies outside the circle.

For point Q(2, 1):
\( S_1 = 2^2 + 1^2 - 4(2) - 6(1) + 9 = 4 + 1 - 8 - 6 + 9 = 0 \)
Since \( S_1 = 0 \), point Q lies on the circle.

For point R(2, 3):
\( S_1 = 2^2 + 3^2 - 4(2) - 6(3) + 9 = 4 + 9 - 8 - 18 + 9 = -4 \)
Since \( S_1 = -4 < 0 \), point R lies inside the circle.
This method provides a quick way to understand the relative position of any point with respect to a given circle.
In simple words: We check each point by putting its x and y values into the circle's equation. If the result is positive, the point is outside. If it's zero, the point is on the circle. If it's negative, the point is inside.

🎯 Exam Tip: Remember the condition for a point \( (x_1, y_1) \) relative to a circle \( S = x^2 + y^2 + 2gx + 2fy + c = 0 \): \( S_1 > 0 \) (outside), \( S_1 = 0 \) (on), \( S_1 < 0 \) (inside).

Question 3. Find the length of the tangent from (1, 2) to the circle \( x^2 + y^2 - 2x + 4y + 9 = 0 \).
Answer: The equation of the circle is \( x^2 + y^2 - 2x + 4y + 9 = 0 \). The point from which the tangent is drawn is \( (x_1, y_1) = (1, 2) \). The length of the tangent (L) from an external point \( (x_1, y_1) \) to the circle \( x^2 + y^2 + 2gx + 2fy + c = 0 \) is given by the formula:
\( L = \sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c} \)
Substituting the coordinates of the point \( (1, 2) \) into the expression \( S_1 = x_1^2 + y_1^2 - 2x_1 + 4y_1 + 9 \):
\( L = \sqrt{1^2 + 2^2 - 2(1) + 4(2) + 9} \)
\( L = \sqrt{1 + 4 - 2 + 8 + 9} \)
\( L = \sqrt{20} \)
\( L = \sqrt{4 \times 5} \)
\( L = 2\sqrt{5} \) units
The length of the tangent is a positive real number, which confirms that the point (1, 2) is outside the circle.
In simple words: To find how long the tangent line is from a point to a circle, we put the point's coordinates into a special square root formula. For point (1, 2) and the given circle, the length of the tangent is \( 2\sqrt{5} \) units.

🎯 Exam Tip: Ensure the given equation of the circle is in the standard form \( x^2 + y^2 + 2gx + 2fy + c = 0 \) before applying the length of tangent formula \( \sqrt{S_1} \).

Question 4. Find the value of P if the line \( 3x + 4y - P = 0 \) is a tangent to the circle \( x^2 + y^2 = 16 \).
Answer: The equation of the line is \( 3x + 4y - P = 0 \). We need to rewrite this in the form \( y = mx + c \):
\( 4y = -3x + P \)
\( y = -\frac{3}{4}x + \frac{P}{4} \)
From this, we get \( m = -\frac{3}{4} \) and \( c = \frac{P}{4} \).
The equation of the circle is \( x^2 + y^2 = 16 \). This is in the form \( x^2 + y^2 = a^2 \), so \( a^2 = 16 \).
The condition for a line \( y = mx + c \) to be a tangent to the circle \( x^2 + y^2 = a^2 \) is \( c^2 = a^2(1 + m^2) \).
Substitute the values of \( c, a^2 \), and \( m \) into the condition:
\( \left(\frac{P}{4}\right)^2 = 16\left(1 + \left(-\frac{3}{4}\right)^2\right) \)
\( \left(\frac{P}{4}\right)^2 = 16\left(1 + \frac{9}{16}\right) \)
\( \frac{P^2}{16} = 16\left(\frac{16+9}{16}\right) \)
\( \frac{P^2}{16} = 16\left(\frac{25}{16}\right) \)
Now, we can multiply both sides by 16:
\( \implies P^2 = 16 \times 25 \)
Take the square root of both sides to find P:
\( \implies P = \pm\sqrt{16 \times 25} \)
\( \implies P = \pm\sqrt{16} \times \sqrt{25} \)
\( \implies P = \pm 4 \times 5 \)
\( \implies P = \pm 20 \)
There can be two possible values of P because a line can be tangent to a circle at two different points for a given slope, or essentially, there are two lines with slope m that are tangent to a given circle.
In simple words: We changed the line equation to show its slope (\( m \)) and y-intercept (\( c \)). We know the circle's radius squared (\( a^2 \)). Using a special rule for when a line touches a circle, we calculated P to be either \( +20 \) or \( -20 \).

🎯 Exam Tip: Always convert the line equation into the slope-intercept form \( y = mx + c \) to correctly identify \( m \) and \( c \). Remember that the value of P can be positive or negative, representing two possible tangent lines.