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The distance formula is an important and well used formula. While using this we need to be careful with negative numbers. There is another method, wherein you draw the diagram as shown below.

You will get a right angled triangle and we need to find the length of the hypotenuse, using Pythagoras's theorem.

Length AB = 4 units and BC = 3 units

Here we have a Pythagorean triplet and therefore the length of the hypotenuse = radius of the circle = 5 .


The Area = \(\pi * r^2 = 25\pi\)


Option D

Arun Kumar
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Using the distance formula, we know that: Distance = sqrt((3-(-1))^2 + (0-(-3))^2))
Coordinates : (3, 0) and (-1, -3)
Distance = 5

Area of circle= pi*r*r
Pi*25

IMO D
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Radius of circle = Distance between Centre (3, 0) and (-1, -3) = sqrt (4^2 + 3^2) = 5

Area of circle = Pi * (Radius^2)= 25*Pi

Answer is D
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Radius = √[(x2-x1)^2+(y2-y1)^2]
Here, x1=3, y1= 0, x2=-1, and y2=-3
r=√[4^2+3^2]
r=√25
r=5

Area of circle = πr^2 = 25π

Option D

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Bunuel
What is the area of a circle if its center is at (3, 0) and the circle passes through (-1, -3)?

A. \(9\pi\)
B. \(10\pi\)
C. \(16\pi\)
D. \(25\pi\)
E. \(36\pi\)


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Alternate way:

Eqn. of a circle
\((x-p)^2 -(y-q)^2=r^2\)

Where \(p\) and \(q\) are the coordinates of the center and \((x,y)\) are the points through which circle passes and \(r\) is the radius of the circle.
\((-1-3)^2 +(-3-0)^2= r^2\)
\(16+9=r^2\) so \(r^2=25\) hence \( r=5 \) and area of circle is \(25\pi\)
Ans-D

Hope it helps.
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