TS EAMCET 9 May 2024 Shift 1 Solved Paper

Three points on circle are
(1,1),(−2,2),(2,−2)
Let equation of circle be
x2+y2+2gx+2fy+c=0
∵ Above points lie on the circle
‌∴‌‌1+1+2g+2f+c=0
‌⇒‌‌2g+2f+c+2=0‌‌...‌ (i) ‌
‌(−2)2+(2)2+2(−2)g+2×2×f+c=0
‌8−4g+4f+c=0
‌‌‌4g−4f−c−8=0‌‌...‌ (ii) ‌
and
(2)2+(−2)2+2×2×g+2(−2)f+c=0
4g−4f+c+8=0‌‌...‌ (iii) ‌
On solving Eqs. (i), (ii) and (iii), we get
‌c=−8,g=‌

3

2

,f=‌

3

2

‌∴‌ Centre of circle ‌≡(‌

−3

2

,‌

−3

2

)
Now, perpendicular distance from centre of the circle to the line
‌=|‌

3(− 3 2 )−4(− 3 2 )+1

√32+(−4)2

|
‌=‌

|− 9 2 + 12 2 +1|

5

=‌

1

2