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Q. Consider the circle `x^2 + y^2 - 10x - 6y + 30=0` . Let O be the centre of the...
Q. Consider the circle `x^2 + y^2 - 10x - 6y + 30=0` . Let O be the centre of the circle and tangent at A(7,3) and B(5,1) meet at C. Let S=0 rpresent the family of circles passing through A and B:-
A) Area of quadrilateral OACB=4
B) Area of quadrilateral OACB=6
C)The co-ordinates of point C are (7,1)
D)The smallest possible circle of the family S=0 is `x^2 + y^2 -12x - 4y + 38=0`
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Best answer as selected by question asker.
Given equation of the circle is `x^2+y^2-10x-6y+30=0`
Therefore, coordinates of O are (5,3) and radius of the circle=2
Differentiating with respect to x,
Equation of tangent at (7,3),
The equation of tangent at (5,1),
Therefore, coordinates of C are (7,1)
Considering quadrilateral OACB, side `BC=sqrt(2^2+0^2)=2`
and, side `AC=sqrt(0^2+2^2)=2`
So, it is a square with side=2, thus, `area= 2^2=4`
Equation of AB is y=x-4, or, `x-y=4`
This is the radical axis.
Equation of smallest circle passing through the points (7,3) and (5,1) is:
Hence, Options A), C) and D) are the correct answers.
Posted by llltkl on July 16, 2013 at 2:16 AM (Answer #1)
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