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Grade: 11
        
Find the equation of the circle which is concentric with the circle x^2 + y ^2 - 2x - 6y - 3= 0 and passes through the point of intersection of the lines 2x + 3y =1 and x - y = 3
2 months ago

Answers : (1)

Aditya Gupta
1819 Points
							
observe that point of intersection of the lines 2x + 3y =1 and x - y = 3 can found out by solving these eqns together, to obtain (2, – 1).
now, x^2 + y ^2 - 2x - 6y - 3= 0 can be written as (x – 1)^2 + (y – 3)^2= 13, so its centre is (1, 3).
as the other circle is conentric with this circle, it too will have the same centre. so its eqn is:
(x – 1)^2 + (y – 3)^2= r^2, where r is its radius. but, it passes through point of intersection of the lines 2x + 3y =1 and x - y = 3, which is (2, – 1).
so, (2 – 1)^2 + ( – 1 – 3)^2= r^2
or r^2= 17
hence eqn becomes
(x – 1)^2 + (y – 3)^2= 17
or x^2 + y^2 – 2x – 6y – 7= 0
kindly approve :))
2 months ago
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