A circle C passing through the point (1,1) bisects the circumference of the circle x2+y2−2x=0. If C is orthogonal to the circle x2+y2+2y−3=0, then the centre of the circle C is
Let the equation of circle C be x2+y2+2gx+2fy+c=0 Since, it passes through (1,1) then 12+12+2g+2f+c=0 ⇒‌‌2+2g+2f+c=0...(i) Now, S1:x2+y2+2gx+2fy+c=0 and S2=x2+y2−2x=0 So, S1−S2=0 is the common chord. ⇒(2g+2x+2fy+c=0 Now, center of s2=x2+y2−2x=0 is (1,0) Since, the common chord is a diameter of S2, then centre of S2 must lies on the common chord. ‌∴‌‌(2g+2(l)+2f(0)+c=0 ‌⇒‌‌2g+2+c=0 Also, circle C is orthogonal to x2+y2+2y−3=0 Here, g1=g,f1=f,c1=c and g2=0, f2=1,c2=−3 Using the orthogonality condition ‌2g1g2+2f1f2=c1+c2 ⇒2g(0)+2f(1)=c+(−3) ⇒2f=c−3 From Eq. (ii), c=−2g−2 Put this in Eq. (iii), we get ⇒‌‌2f‌=c−3=−2g−2−3\⇒‌‌2f‌=−2g−5 Substitute c=−2g−2 into Eq. (i), we get ‌⇒2+2g+2f−2g−2=0 ‌⇒2f=0 ‌⇒f=0
Put f=0 in Eq. (iv), we get ‌2(0)‌=−2g−5 ⇒‌2g‌=−5 ⇒‌g‌=‌
−5
2
The center of circle C is ‌(−g,−f)=(−(−‌