A circle S given by x2+y2−14x+6y+33=0 cuts the X-axis at A and B(OB>OA).C is mid-point of AB.L is a line through C and having slope ( -1 ). If L is the diameter of a circle S′ and also the radical axis of the circles S and S′, then the equation of the circle S′ is
Given equation of circle x2+y2−14x+6y+33=0...‌ (i) ‌
‌ Centre of circle ‌=‌(−g,−f) =‌(7,−3) ‌ and radius of circle ‌‌=√g2+f2−c ‌=√49+9−33 ‌=√25=5 Since, circle cuts theX-axis,y=0 ‌x2−14x+33=0 ⇒x2−11x−3x+33=0 ⇒x(x−11)−3(x−11)=0 ⇒(x−3)(x−11)=0 ⇒x=3,11 ∴‌A=(3,0)‌ and ‌B=(11,0) Since,Cis the mid-point ofAB ∴‌‌C‌=(‌
3+11
2
,‌
0+0
2
) ‌=(7,0) ∴Equation of lineL, which passes through C and having slope(−1)is ‌y−0=(−1)(x−7) ‌⇒‌‌y=−x+7 ‌⇒‌‌x+y−7=0...‌ (ii) ‌ Since, L is the radical axis of the circlessandS′(S−S′=0) ‌∴‌‌(x2+y2−14x+6y+33)−(x2+y2+2g′x+2f′y+c′)=0 ‌⇒(−14−2g′)x+(6−2f′)y+(33−c′)=0 Which is identical with line (ii) ∴‌‌
−14−2g′
1
=‌
6−2f′
1
=‌
33−c′
−7
=k ⇒−14−2g′=k‌ and ‌6−2f′=k ∴‌−14−2g′=6−2f′ ⇒g′−f′=−10...‌ (iii) ‌ Since, centre ofS′lies on (ii) ∴‌−g′−f′‌=7"‌ ⇒‌g′+f′‌=−7........(iv) Solving Eqs. (iii) and (iv), we get ‌g′=‌
−17
2
,f′=‌
3
2
∴‌‌−14−2(−‌
17
2
)=k ⇒‌‌k=3 Also, ‌
33−c′
−7
=k=3 ‌⇒‌‌33−c′=−21 ‌⇒‌‌c′=54 ∴ Equation of required circle is ‌x2+y2+2(‌