Find the equation of the circle having `(1,-2)`as its centre and passing through the intersectionof the lines `3x+y=14a d n2x+5y=18.`

Find the equation of the circle having `(1,-2)`as its centre and passi

1.	Find the equation of the circle having `(1,-2)`as its centre and passing through the intersectionof the lines `3x+y=14a d n2x+5y=18.`
Answer» Given that, centre of the circle is (1,-2) and the circle passing through the lines 3x+y= 14….(i) and 2x+5y=18…(ii) From Eq. (i) y=14-3x pu in Eq. (ii), we get 2x+70-15x=18 `rArr` -13x=-52`rArr`x=4 Now, x=4 put in Eq. (i) we get 12+y=14`rArr`=y=2 Since, point (4, 2) lie on these lines also lies on the circle. Radius of the circle =`sqrt((4-1)^(2)+(2+2)^(2))` `=sqrt(9+16)=5` Now equation of the circle is `(x-1)^(2)+(y+2)^(2)=5^(2)` `rArr x^(2)-2x+1+y^(2)+4y+4=25` `rArr x^(2)+y^(2)-2x+4y-20=0`

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