JEE Mathematics 2026 Question with Solution

Given: $|z-6|=5$ and $|z+2-6i|=5$.

Find: The value of $z^3+3z^2-15z+14$.

Concept: The given conditions represent two circles in the Argand plane. If two circles with equal radii touch externally, the common point is the midpoint of their centres.

Step 1: Identify the centres of the circles.

$$|z-6|=5 \Rightarrow \text{Center } C_1=(6,0)$$ $$|z+2-6i|=5 \Rightarrow \text{Center } C_2=(-2,6)$$

Distance between centres:

$$\sqrt{(6+2)^2+(0-6)^2}=\sqrt{64+36}=10$$

Since the distance between centres equals the sum of radii, the circles touch externally.

Step 2: Find the point of contact.

$$z=\left(\frac{6+(-2)}{2},\frac{0+6}{2}\right)=(2,3)$$

Hence,

$$z=2+3i$$

Step 3: Evaluate the expression.

$$z^2=(2+3i)^2=-5+12i$$ $$z^3=(2+3i)(-5+12i)=-46+9i$$

Now,

$$z^3+3z^2-15z+14=(-46+9i)+3(-5+12i)-15(2+3i)+14$$ $$=(-46-15-30+14)+(9+36-45)i=37$$

Therefore, the value of the expression is $37$. The correct option is A.

Given: Both loci are circles of radius $5$.

Find: The value of $z^3+3z^2-15z+14$.

The centres are $(6,0)$ and $(-2,6)$. Their distance is

$$\sqrt{8^2+(-6)^2}=10$$

which is exactly the sum of the radii. So the circles touch externally, and the common point is the midpoint of the two centres.

$$z=\left(\frac{6+(-2)}{2},\frac{0+6}{2}\right)=(2,3)$$

Thus $z=2+3i$. Substituting this into the expression gives $37$.

Therefore, the correct option is A.