Differentiation (Part-38) | S N De

Tangent and Normal | Part-4

In the previous article , we have solved few VSA type Questions. In the following article, we are going to discuss/solve few more Very Short Answer Type Questions of S.N.Dey Mathematics-Class 12 of the chapter Tangent and Normal (Ex-14).

Find the lengths of the tangents drawn from the point $~[27-30]:$

27. $(x_1,y_1)~$ to the circle $~x^2+y^2+2x=0$

Solution.

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

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

$\sqrt{x_1^2+y_1^2+2x_1}~~ \text{unit}$

28. $(-4,5)~$ to the circle $~x^2+y^2=16.$

Solution.

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

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

$\sqrt{(-4)^2+5^2-16}=\sqrt{16+25-16}=5~~\text{unit}$ 

29. $(-1,1)~$ to the circle $~x^2+y^2-2x-4y+1=0.$

Solution.

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

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

$\sqrt{(-1)^2+1^2-2(-1)-4 \times 1+1}\\=\sqrt{1+1+2-4+1}=1~~ \text{unit}$

30. $(2,-2)~$ to the circle  $~3(x^2+y^2)-4x-7y=3.$

Solution.

$3(x^2+y^2)-4x-7y-3=0\\ \text{or,}~~ x^2+y^2-\frac 43~x-\frac 73~y-1=0 \longrightarrow(1)$

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

$\sqrt{2^2+(-2)^2-\frac 43 \times 2-\frac 73  \times (-2)-1}\\=\sqrt{4+4-\frac 83+\frac{14}{3}-1}\\=\sqrt{7+\frac{14-8}{3}}\\=\sqrt{7+\frac 63}=\sqrt{7+2}=3~~ \text{unit}$

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.

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

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

Comparing (1) and (2) with $~x^2+y^2+2gx+2fy+c=0~$ we get,

$~2g=-4 \Rightarrow g=-2,~~ 2f=6 \Rightarrow f=3.$

Clearly, the circles (1) and (2) are concentric and the centre is given  by $~(-g,-f)=(2,-3).$

The radius of circle (1) is 

$r_1=\sqrt{g^2+f^2-c}=\sqrt{(-2)^2+3^2-(-2)}\\~~=\sqrt{15}~~\text{unit}$

The radius of circle (2) is 

$r_2=\sqrt{g^2+f^2-c}=\sqrt{(-2)^2+3^2-7}\\~~=\sqrt{6}~~\text{unit}$

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

$\sqrt{r_1^2-r_2^2}=\sqrt{15-6}=\sqrt{9}=3~~\text{unit}$