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Question

A foot of the normal from the point (4,3) to a circle is (2,1) and a diameter of the circle has the equation 2x-y=2. Then the equation of the circle is


A

x2+y2+2x-1=0

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B

x2+y2-2x-1=0

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C

x2+y2-2y-1=0

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D

None of these

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Solution

The correct option is B

x2+y2-2x-1=0


The explanation for the correct answer:

Step 1: Find the equation of the line:

Given: The normal from the point (x1,y1) is (4,3) to a circle (x2,y2)is (2,1) and

The diameter of the circle is 2x-y=2

The line equation that joints these two points is y-y1=m(x-x1)

where m is the slope of the given line.

Slope, m = y2-y1x2-x1

Thus the equation of the line is y-3=1-32-4(x-2)

⇒ y-3=1(x-4)

⇒ x-y=1…….[1]

The diameter of the circle is 2x-y=2……..[2]

The point of intersection of these lines gives the center of the circle.

By subtracting the equations [1] and [2], we get x=1and y=0

Thus the equation of the circle can be obtained from the center (x1,y1) is (1,0) and the point (x2,y2)is (2,1).

Step 2: Find the equation of the circle

The radius of the circle is given by r=(x2-x1)2+(y2-y1)2

=(2-1)2+(1-0)2

=1+1

=2

The equation of the circle is given as (x-x1)2+(y-y1)2=r2

⇒ (x-1)2+(y-0)2=(2)2

⇒ x2-2x+1+y2=2

⇒ x2+y2-2x-1=0

Hence, option (B) is the correct option.


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