AP EAPCET 05-Jul-2022 Shift 1 Paper

Given equation (x2+y2−4x−2y−20=0)
So, C=(2,1) and radius =√(g)2+(f)2−c
‌=√(2)2+(1)2+20
When substituted x=10 and y=7 in equation, then value becomes
which is greater than zero.
Thus, the point (10,7) lies outside the circle.
Its distance from the centre of the circle (2,1) is √(10−42+(7−1)2.=10 units
So, the minimum distance from the circle therefore becomes 10−5=5 units