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Bunuel
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In the rectangular coordinate system, points (4, 0) and 28 Aug 2023, 02:58
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hrithiksindwani
Where am I going wrong with this:

If the length of chord is 8.
The radius must be joining both end of the Chord and forming a triangle where the two opposite radius will give angles 45, 45, and thus angle at the centre becomes 90. This makes this a 1:1:Root 2 triangle and hence Radius 4 root 2.

Can you please help?
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The red part is not correct. Why do you assume that?
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In the rectangular coordinate system, points (4, 0) and 28 Aug 2023, 03:30
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Given: In the rectangular coordinate system, points (4, 0) and (– 4, 0) both lie on circle C.
Asked: What is the maximum possible value of the radius of C ?

The is no finite maximum value of radius C since these 2 point (4,0) & (-4,0) can form an arc of a infinitely large circle C.

IMO E
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In the rectangular coordinate system, points (4, 0) and 28 Aug 2023, 06:36
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hrithiksindwani
Where am I going wrong with this:

If the length of chord is 8.
The radius must be joining both end of the Chord and forming a triangle where the two opposite radius will give angles 45, 45, and thus angle at the centre becomes 90. This makes this a 1:1:Root 2 triangle and hence Radius 4 root 2.

Can you please help?
Bunuel
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The opposite angles will be equal but how do you figure that they will be 45 - 45 each? They could be 10 degrees each or 80 degrees each etc.
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In the rectangular coordinate system, points (4, 0) and 16 May 2026, 23:11
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Think visually, not formula-first.

You have two points:

(-4,0) and (4,0)

These are 8 units apart.

So imagine two pins fixed on a paper.

Now ask:

“How small can a circle be while still passing through both pins?”

The smallest possible circle happens when the two points sit exactly opposite each other on the circle.

That is called a diameter.

Like this:

(-4,0) -------- center -------- (4,0)

Distance between endpoints = 8

Diameter = 8

Radius is half the diameter:

8/2 = 4

So:

• radius CANNOT be smaller than 4
• but it CAN be larger

Why larger?

Because the center does NOT have to stay in the middle line.

You can move the center upward.

Example idea:

center

.-------------.
/ \
(-4,0) (4,0)

Now the circle becomes bigger.

Move center even farther away:

center


.-------------------.
/ \
(-4,0) (4,0)

Even bigger radius.

So:

• minimum radius = 4
• no maximum radius

Answer = E

The important concept is:

A fixed chord does NOT determine one unique circle.
Many circles can pass through the same two points.



The only thing we can conclude from the question is that center lies on the y-axis. But it could be ANY point on it, hence we cannot determine maximum value of r.

How was it concluded that center lies on y-axis?

Logic:

The two points are:

(4,0) and (-4,0)

Notice something important:

They are mirror images across the y-axis.

Now use a circle fact:

The center of a circle is always on the perpendicular bisector of any chord.

The segment joining the two points is:

(-4,0) ---------------- (4,0)

This segment:

• is horizontal
• has midpoint at:

((4 + -4)/2 , (0 + 0)/2) = (0,0)

So its perpendicular bisector is:

• the vertical line through (0,0)
• which is the y-axis:

x = 0

Therefore the center must lie somewhere on the y-axis.

Why?

Because the center must be equally far from both points.

Any point on the y-axis is symmetric relative to these two points, so distances match.

For example center at:

(0,5)

Distance to both points:

√(42 + 52)

same for each point.

So:

• every possible center lies on y-axis
• but it can move infinitely far up/down
• hence radius can become infinitely large

Therefore:

No finite maximum.
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