Students can download 11th Business Maths Chapter 3 Analytical Geometry Ex 3.5 Questions and Answers, Notes, Samcheer Kalvi 11th Business Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

## Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 3 Analytical Geometry Ex 3.5

### Samacheer Kalvi 11th Business Maths Analytical Geometry Ex 3.5 Text Book Back Questions and Answers

Question 1.

Find the equation of the tangent to the circle x^{2} + y^{2} – 4x + 4y – 8 = 0 at (-2, -2).

Solution:

The equation of the tangent to the circle x^{2} + y^{2} – 4x + 4y – 8 = 0 at (x_{1}, y_{1}) is

xx_{1} + yy_{1} – 4\(\frac{\left(x+x_{1}\right)}{2}\) + 4\(\frac{\left(y+y_{1}\right)}{2}\) – 8 = 0

Here (x_{1}, y_{1}) = (-2, -2)

⇒ x(-2) + y(-2) – 2(x – 2) + 2(y – 2) – 8 = 0

⇒ -2x – 2y – 2x + 4 + 2y – 4 – 8 = 0

⇒ -4x – 8 = 0

⇒ x + 2 = 0

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.

Solution:

The equation of the circle is x^{2} + y^{2} – 4x – 6y + 9 = 0

PT^{2} = \(x_{1}^{2}+y_{1}^{2}\) – 4x_{1} – 6y_{1} + 9

At P(1, 0), PT^{2} = 1 + 0 – 4 – 0 + 9 = 6 > 0

At Q(2, 1), PT^{2} = 4 + 1 – 8 – 6 + 9 = 0

At R(2, 3), PT^{2} = 4 + 9 – 8 – 18 + 9 = -4 < 0

The point P lies outside the circle.

The point Q lies on the circle.

The point R lies inside the circle.

Question 3.

Find the length of the tangent from (1, 2) to the circle x^{2} + y^{2} – 2x + 4y + 9 = 0.

Solution:

The length of the tangent from (x_{1}, y_{1}) to the circle x^{2} + y^{2} – 2x + 4y + 9 = 0 is \(\sqrt{x_{1}^{2}+y_{1}^{2}-2 x_{1}+4 y_{1}+9}\)

Length of the tangent from (1, 2) = \(\sqrt{1^{2}+2^{2}-2(1)+4(2)+9}\)

= \(\sqrt{1+4-2+8+9}\)

= \(\sqrt{20}\)

= \(\sqrt{4 \times 5}\)

= 2√5 units

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.

Solution:

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})

Equation of the line is 3x + 4y – P = 0

Equation of the circle is x^{2} + y^{2} = 16

4y = -3x + P

y = \(\frac{-3}{4} x+\frac{P}{4}\)

∴ m = \(\frac{-3}{4}\), c = \(\frac{P}{4}\)

Equation of the circle is x^{2} + y^{2} = 16

∴ a^{2} = 16

Condition for tangency we have c^{2} = a^{2}(1 + m^{2})

⇒ \(\left(\frac{P}{4}\right)^{2}=16\left(1+\frac{9}{16}\right)\)

⇒ \(\frac{P^{2}}{16}=16\left(\frac{25}{16}\right)\)

⇒ P^{2} = 16 × 25

⇒ P = ±√16√25

⇒ P = ±4 × 5

⇒ P = ±20