Find the equation of a circle which passes through the point `(2,0)` and whose centre is the limit of the point of intersection of eth lines `3x+5y=1, and (2+c)x+5c^2y=1, c to1.`

Answer

`25(x^(2)+y^(2))-20x+2y-60=0`

Explanation

Given lines are
`3x + 5y -1 = 0" "...(i)`
and `(2+c)x+5c^(2)y-1=0" "...(ii)`
`therefore (x)/(-5+5c^(2))=(y)/(-(2+c)+3)=(1)/(15c^(2)-5c-10)`
`rArr x = (5(c^(2)-1))/(5(3c^(2)-c-2))and y = (1)/(15c^(2)-5c-10)`
`rArr underset(cto1)limx=underset(cto1)lim(2c)/(6c-1)`
and ` underset(cto1)limy=underset(cto1)lim(-1)/(30c-6)rArr underset(cto1)limx=(2)/(5)`
`and underset(cto1)limy=(1)/(25)`
`therefore " Centre " (underset(cto1)limx,underset(cto1)limy)=((2)/(5),-(1)/(25))`
`therefore " Radius " = sqrt((2-(2)/(5))^(2)+(0+(1)/(25))^(2))=sqrt((64)/(25)+(1)/(625))=(sqrt1601)/(25)`
`therefore` Equation of the circle is `(x-(2)/(5))^(2)+(y+(1)/(25))^(2)=(1601)/(625)`
`rArr x^(2)+y^(2)-(4x)/(5)+(2y)/(25)+(4)/(25)+(1)/(625)-(1601)/(625)=0`
`rArr 25(x^(2)+y^(2))-20x+2y-60=0`