TS EAMCET 20-July-2022 Shift 2 Solved Paper

Let the equation of circle whose centre is (h,k)=
‌(x−h)2+(y−k)2=r2
⇒x2+h2−2xh+y2+k2−2ky=r2
⇒x2+y2−2xh−2ky=r2−h2−k2
⇒x2+y2−2xh−2ky+c=0
(Let h2+k2−r2=c)
Thus also pass origin then,
‌c‌=0
∴‌‌x2+y2−2xh−2ky=0 ....(i)
(i) is orthogonal to x2+y2+2x+6y+12=0 and x2+y2+4x−6y+9=0
then −4h+6k=9.....(ii)
and −2h−6k=12 ...(iii)
Adding Eqs. (ii) and (iii), we get
−6h‌=21
h‌=−‌

21

6

=‌

−7

2

Putting the value of h in Eq. (ii), we get
k=−‌

5

6

Now,