The centres of all circles passing through the points of intersection of the circles x2+y2+2x−2y+1=0 and x2+y2−2x+2y−2=0 and having radius √14 lie on the curve
Given, equation of circles ‌x2+y2+2x−2y+1=0‌ and ‌ ‌x2+y2−2x+2y−2=0 ∴ Equation of circle passing through the point of intersection of two given circle is (x2+y2+2x−2y+1)+λ(x2+y2−2x+2y−2)=0 ⇒‌‌(1+λ)x2+(1+λ)y2+(2−2λ)x+(−2+2λ)y+(1−2λ)=0 ⇒‌‌x2+y2+‌
(2−2λ)
1+λ
x+‌
(−2+2λ)
1+λ
y+‌
1−2λ
1+λ
=0 Coordinate of centre of circle =(−g,−f)=(−‌
1−λ
1+λ
,−‌
−1+λ
1+λ
) ∴ Radius of ‌(−‌
1−λ
1+λ
)2+(−‌
−1+λ
1+λ
)2−(‌
1−2λ
1+λ
)=14 ‌⇒‌‌‌
(1−λ)2
(1+λ)2
+‌
(λ−1)2
(1+λ)2
−‌
1−2λ
1+λ
=14 ‌⇒‌‌(1−λ)2+(λ−1)2−(1+λ)(1−2λ) ‌=14(1+λ)2 ‌⇒‌‌2(λ−1)2−(1−2λ+λ−2λ2) ‌=14(1+λ2+2λ) ‌⇒‌‌2(λ2+1−2λ−(1−λ−2λ2). ‌=14(1+λ2+2λ) ‌⇒‌‌4λ2+1−3λ=14+14λ2+28λ ‌⇒‌‌10λ2+31λ+13=0 ⇒ This is a quadrate equation is λ the existence of real values for λ conforms that such circle exists. Coordinates of centre ‌h=‌