AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper

Consider the diagram, By the property of distances,
|4h−3k−245|=|

4h+3k−42

5

|
=√(h−2)2+(k−8)2
This implies,
4h−3k−24=4h+3k−42
6k=18
k=3
Or,
4h−3k−24=−4h−3k+42
8h=66
h=

66

8

=

33

4

>8
Now,
(4h−3k−24)2=25[(h−2)2 +(k−8)2]
From above,
h = 2
Thus,the centre is ( 2,3) .
The required circle is,
(x−2)2+(y−3)2=25
x2+4−4x+y2+9−6y=25
x2+y2−4x−6x=12