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`        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`
5 months ago

```							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^2or r^2= 17hence eqn becomes(x – 1)^2 + (y – 3)^2= 17or x^2 + y^2 – 2x – 6y – 7= 0kindly approve :))
```
5 months ago
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