NIMCET 2021 — Mathematics PYQ

1. Identify parameters for Circle 1:

The equation is in the form $x^2+y^2=r^2$.

• Center $C_1=(0,0)$

• Radius $r_1=\sqrt{4}=2$

2. Identify parameters for Circle 2:

The equation is $x^2+y^2-6x-8y-24=0$.

Comparing with the general form $x^2+y^2+2gx+2fy+c=0$:

• Center $C_2=(-g,-f)=(3,4)$

• Radius $r_2=\sqrt{g^2+f^2-c}=\sqrt{(-3{)}^2+(-4{)}^2-(-24)}=\sqrt{9+16+24}=\sqrt{49}=7$

3. Calculate the distance between centers ($d$):

Using the distance formula between $C_1(0,0)$ and $C_2(3,4)$:

$$d=\sqrt{(3-0{)}^2+(4-0{)}^2}=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$$

4. Compare distance ($d$) with radii sum ($r_1+r_2$) and difference ($|r_1-r_2|$):

• Sum of radii: $r_1+r_2=2+7=9$

• Difference of radii: $|r_2-r_1|=|7-2|=5$

Since the distance between the centers $d=5$ is exactly equal to the difference of their radii $|r_2-r_1|=5$, the two circles touch each other internally.

When two circles touch internally, there is exactly 1 common tangent at the point of contact.

Final Answer

The correct option is B (1).