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circle x^2 + y^2 = 1 has radius 1.

from the origin(center of circle) to point (3,4), it forms a right angle triangle. Then we find hypotenuse side (min. dist. to the point). which is 5.

5 - 1 = 4.
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Why is it assumed that the centre of the circle is at origin?
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Why is it assumed that the centre of the circle is at origin?

Circle on a plane

In an x-y Cartesian coordinate system, the circle with center (a, b) and radius r is the set of all points (x, y) such that:
\((x-a)^2+(y-b)^2=r^2\)




This equation of the circle follows from the Pythagorean theorem applied to any point on the circle: as shown in the diagram above, the radius is the hypotenuse of a right-angled triangle whose other sides are of length x-a and y-b.

If the circle is centered at the origin (0, 0), then the equation simplifies to:
\(x^2+y^2=r^2\)

Therefore, x^2 + y^2 = 1, represents a circle centered at the origin and the radius of \(\sqrt{1}=1\).

24. Coordinate Geometry



For more check:
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Hope it helps.
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GMAT CLUB should take down these questions of co-ordinate and Geometry questions.

Not relevant anyway

DIstance of (3, 4) from center of circle (origin) = √(3^2 + 4^2) = 5

Radius = 1

Min. distance = 5-1 = 4

Answer: Option D
MathRevolution
What is the minimum distance between the point \((3,4)\)and points on the circle \(x^2+y^2=1\)?

A. 1
B. 2
C. 3
D. 4
E. 5
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