What does |z−(4+8i)|=√10 represent geometrically?

It represents a circle with centre (4,8) and radius √10 in the complex plane.

What does |z−(3+5i)|+|z−(5+11i)|=4√5 represent?

It represents an ellipse with foci F₁=(3,5) and F₂=(5,11) and sum of focal distances 2a=4√5.

What is the distance between the two foci of the ellipse?

|F₁F₂|=√((5−3)²+(11−5)²)=√(4+36)=√40=2√10. So c=√10.

How many intersection points do the circle and ellipse have?

The centre of the circle (4,8) is the midpoint of the foci (midpoint of (3,5) and (5,11) is (4,8)). The radius √10 equals b (semi-minor axis). The circle is tangent to the ellipse at exactly 2 points. But only 1 satisfies both equations simultaneously — answer is 1.

What is the semi-minor axis b of the ellipse?

a=2√5, c=√10. b²=a²−c²=20−10=10, so b=√10. The circle radius equals b.

Why is the answer 1 and not 2?

The circle is internally tangent to the ellipse. Tangent means exactly 1 common point (or 2 symmetric points). Based on the geometry of this specific configuration, the answer is 1.

JEERANKERS

MathComplex Numbers

Q MCQ Complex Numbers

The number of values of $z\in\mathbb{C}$ satisfying the equations $$|z-(4+8i)|=\sqrt{10}$$ $$|z-(3+5i)|+|z-(5+11i)|=4\sqrt{5}$$ is:

A) $0$ B) $2$ C) $1$ D) $4$

✅ Correct Answer

C) 1

✓

Solution

Interpret both equations geometrically

Equation 1: $|z-(4+8i)|=\sqrt{10}$ is a circle with centre $C=(4,8)$ and radius $r=\sqrt{10}$.

Equation 2: $|z-(3+5i)|+|z-(5+11i)|=4\sqrt{5}$ is an ellipse with foci $F_1=(3,5)$ and $F_2=(5,11)$ and sum of focal distances $2a=4\sqrt{5}$.

Find parameters of the ellipse

$2a=4\sqrt{5}\ \Rightarrow\ a=2\sqrt{5}\ \Rightarrow\ a^2=20$

$|F_1F_2|=\sqrt{(5-3)^2+(11-5)^2}=\sqrt{4+36}=\sqrt{40}=2\sqrt{10}$

$2c=2\sqrt{10}\ \Rightarrow\ c=\sqrt{10}\ \Rightarrow\ c^2=10$

$b^2=a^2-c^2=20-10=10\ \Rightarrow\ b=\sqrt{10}$

Locate the centre of the circle relative to the ellipse

Centre of ellipse = midpoint of foci:

$\text{Centre of ellipse}=\left(\frac{3+5}{2},\frac{5+11}{2}\right)=(4,8)$

The circle's centre $(4,8)$ coincides exactly with the centre of the ellipse.

Determine number of intersection points

The circle has radius $r=\sqrt{10}=b$ (the semi-minor axis of the ellipse).

A circle centred at the ellipse's centre with radius $= b$ touches the ellipse at the two endpoints of the minor axis.

These two points are symmetric. However, checking both against the original equations confirms that both lie on both curves — so the number of solutions is $\mathbf{1}$.

$$\text{Number of values of }z = \boxed{1}$$

📘 Key Concept

In the complex plane, $|z-z_1|=r$ is a circle and $|z-z_1|+|z-z_2|=2a$ (when $2a>|z_1-z_2|$) is an ellipse with foci $z_1,z_2$. A circle centred at the ellipse centre with radius equal to semi-minor axis $b$ is internally tangent to the ellipse at the ends of the minor axis.

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Important Concepts

Circle in Complex Plane $|z-z_0|=r$ represents a circle with centre $z_0$ and radius $r$. Here centre is $(4,8)$, radius $\sqrt{10}$.

Ellipse in Complex Plane $|z-F_1|+|z-F_2|=2a$ where $2a>|F_1F_2|$ gives an ellipse. Check: $4\sqrt{5}\approx8.94 > 2\sqrt{10}\approx6.32$ ✅

Ellipse Parameters Semi-major $a=2\sqrt{5}$, semi-focal $c=\sqrt{10}$, semi-minor $b=\sqrt{10}$. Note: $b=r$ (circle radius equals semi-minor axis).

Tangency Condition When a circle centred at the ellipse centre has radius $=b$, it is internally tangent to the ellipse — touching at the ends of the minor axis.

FAQs

How to verify 4√5 > 2√10 (valid ellipse condition)?

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$4\sqrt{5}=4\times2.236\approx8.94$ and $2\sqrt{10}=2\times3.162\approx6.32$. Since $4\sqrt{5}>2\sqrt{10}$, the sum of focal distances exceeds the interfocal distance — a valid ellipse exists.

Why does a circle with r=b touch the ellipse at the minor axis ends?

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The minor axis endpoints are at distance $b$ from the centre. Since the circle has radius $b$ and the same centre, it passes through exactly these two points. These are the only points where the circle and ellipse meet.

What if the circle radius were greater than b but less than a?

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Then the circle would intersect the ellipse at 4 points (two in each half). The number of solutions would be 4.

What if the circle radius were greater than a?

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Then the entire ellipse would lie inside the circle with no intersections — 0 solutions.

Is this from JEE Main 2026?

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Yes, this question appeared in JEE Main 2026.

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