(2+c)x+5c2(51−3x)=1x=2+c−3c21−c2,y=51−3x=5(2+c−3c2)c−1h=c→1lim(1−c)(2+3c)(1−c)(1+c)=52K=c→1lim−5(c−1)(3c+2)c−1=−251Centre(252,−251) r=(2−52)2+(0−251)2=2564+6251r=25161(x−52)2+(y+251)2=125161⇒25x2+25y2−20x+2y−60=0
Let a circle passing through (2,0) have its centre at the point (h,k). Let (xc,yc) be the point of intersection of the lines 3x+5y=1 and (2+c)x+5c2y=1. If h=c→1limxc and k=c→1limyc, then the equation of the circle is :
Held on 9 Apr 2024 · Verified 6 Jul 2026.
25x2+25y2−2x+2y−60=0
5x2+5y2−4x+2y−12=0
5x2+5y2−4x−2y−12=0
25x2+25y2−20x+2y−60=0
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