Questions
The equation of the smallest circle passing from points and and always in the first quadrant, is
a
x2+y2−4x−2y+4=0
b
x2+y2+2x+4y+4=0
c
x2+y2−3x−3y+4=0
d
x2+y2−5x−y+4=0
detailed solution
Correct option is C
Let the equation of the required circle bex2+y2+2gx+2fy+c=0It passes through (1, 1) and (2, 2).∴ 2+2g+2f+c=08+4g+4f+c=0From (ii) and (iii), we getb+f=−3 and c=4Let r be the radius of the circle. Then,r2=g2+f2−c⇒r2=g2+f2−4⇒r2=g2+(3+g)2−4⇒r2=2g2+6g+5 [∵g+f=−3]⇒ r2=2g2+3g+52⇒r2=2g+322+14Clearly, r is least when g+32=0 i.e. g=−32Substituting the values of g, f and c in (i), we getx2+y2−3x−3y+4=0Talk to our academic expert!
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