Differentiation (Part-38) | S N De

Tangent and Normal | Part-4

In the previous article, we solved several VSA type questions. In this article, we continue discussing step-by-step solutions for Very Short Answer Type Questions from S.N. Dey Mathematics (Class 12), Chapter: Tangent and Normal (Ex-14).

Find the lengths of the tangents drawn from the specified point [27–30]:

Question 27

Find the length of the tangent from $(x_1,y_1)$ to the circle $x^2 + y^2 + 2x = 0$.

Solution

$$x^2 + y^2 + 2x = 0 \quad \longrightarrow(1)$$

The length of the tangent from the point $(x_1,y_1)$ to the circle (1) is:

$$\sqrt{x_1^2 + y_1^2 + 2x_1} \quad \text{unit} \quad \mathbf{\color{#059669}{(Ans)}}$$

Question 28

Find the length of the tangent from $(-4,5)$ to the circle $x^2 + y^2 = 16$.

Solution

$$x^2 + y^2 = 16 \implies x^2 + y^2 - 16 = 0 \quad \longrightarrow(1)$$

The length of the tangent from the point $(-4,5)$ to the circle (1) is:

$$\sqrt{(-4)^2 + 5^2 - 16} = \sqrt{16 + 25 - 16} = \sqrt{25} = 5 \quad \text{units} \quad \mathbf{\color{#059669}{(Ans)}}$$

Question 29

Find the length of the tangent from $(-1,1)$ to the circle $x^2 + y^2 - 2x - 4y + 1 = 0$.

Solution

$$x^2 + y^2 - 2x - 4y + 1 = 0 \quad \longrightarrow(1)$$

The length of the tangent from the point $(-1,1)$ to the circle (1) is:

$$\sqrt{(-1)^2 + 1^2 - 2(-1) - 4(1) + 1} = \sqrt{1 + 1 + 2 - 4 + 1} = \sqrt{1} = 1 \quad \text{unit} \quad \mathbf{\color{#059669}{(Ans)}}$$

Question 30

Find the length of the tangent from $(2,-2)$ to the circle $3(x^2 + y^2) - 4x - 7y = 3$.

Solution

$$3(x^2 + y^2) - 4x - 7y - 3 = 0 \implies x^2 + y^2 - \frac{4}{3}x - \frac{7}{3}y - 1 = 0 \quad \longrightarrow(1)$$

The length of the tangent from the point $(2,-2)$ to the circle (1) is:

$$\sqrt{2^2 + (-2)^2 - \frac{4}{3}(2) - \frac{7}{3}(-2) - 1} = \sqrt{4 + 4 - \frac{8}{3} + \frac{14}{3} - 1}$$ $$= \sqrt{7 + \frac{14 - 8}{3}} = \sqrt{7 + \frac{6}{3}} = \sqrt{7 + 2} = \sqrt{9} = 3 \quad \text{units} \quad \mathbf{\color{#059669}{(Ans)}}$$

Question 31

Find the length of the tangent from any point on the circle $x^2 + y^2 - 4x + 6y - 2 = 0$ to the circle $x^2 + y^2 - 4x + 6y + 7 = 0$.

Solution

$$x^2 + y^2 - 4x + 6y - 2 = 0 \quad \longrightarrow(1)$$ $$x^2 + y^2 - 4x + 6y + 7 = 0 \quad \longrightarrow(2)$$

Comparing (1) and (2) with the standard circle equation $x^2 + y^2 + 2gx + 2fy + c = 0$:

$$2g = -4 \implies g = -2 \quad \text{and} \quad 2f = 6 \implies f = 3$$

Clearly, circles (1) and (2) are concentric with a common center at $(-g, -f) = (2, -3)$.

The radius $r_1$ of circle (1) is:

$$r_1 = \sqrt{g^2 + f^2 - c_1} = \sqrt{(-2)^2 + 3^2 - (-2)} = \sqrt{4 + 9 + 2} = \sqrt{15} \quad \text{units}$$

The radius $r_2$ of circle (2) is:

$$r_2 = \sqrt{g^2 + f^2 - c_2} = \sqrt{(-2)^2 + 3^2 - 7} = \sqrt{4 + 9 - 7} = \sqrt{6} \quad \text{units}$$

Therefore, the length of the tangent from any point on circle (1) to circle (2) is:

$$\sqrt{r_1^2 - r_2^2} = \sqrt{(\sqrt{15})^2 - (\sqrt{6})^2} = \sqrt{15 - 6} = \sqrt{9} = 3 \quad \text{units} \quad \mathbf{\color{#059669}{(Ans)}}$$