AP EAMCET Engineering 22 Sep 2020 Shift 2 Paper

Given circle is x2+y2−6x−4y−12=0
Centre =(3,2)
Radius =√(3)2+(2)2+12=√9+4+12=√25
r=5
Since, tangents are parallel to X-axis
∴ Let y=k be the tangent to the given circle
⇒y−k=0 touches the circle
∴ Perpendicular distance
from C(3,2) to y−k=0= radius
‌

|2−k|

√1

=5
|2−k|=5
2−k=±5
2−k=5 (or ) 2−k=−5
k=−3 (ог) k=7
∴ Required equation of tangents are
y=−3 and y=7
⇒(y+3)=0 and (y−7)=0
∴ Equation of pair of tangents is (y+3)(y−7)=0
y2−4y−21=0