x2+y2+ax+2ay+c=0(a<0) For a circle in standard form x2+y2+2gx+2fy+c=0 x-intercept =2√g2−c Here, 2g=a⇒g=a∕2 2f=2a⇒f=a x-intercept ⇒2√‌
a2
4
−c=2√2 ⇒‌‌‌
a2
4
−c=2.....(i) y-intercept ⇒2√a2−c=2√5 a2−c=5...(ii) Subtracting Eq. (ii) from Eq. (i), we get (‌
3
4
)a2=3 ⇒a2=4⇒a=±2 As, a<0, so, a=−2,c=−1 So, x2+y2−2x−4y−1=0 (x−1)2+(y−2)2−1=1+4 ⇒‌‌(x−1)2+(y−2)2=6 Centre (1,2), Radius =√6 Equation of tangent is perpendicular to x+2y=0. So, equation of tangent will be 2x−y+λ=0 Now, perpendicular distance of this line from (1,2) will be √6 units. ‌
|2.1−2+λ|
√5
=√6 λ=±√30 So, T⇒2x−y±√30=0 Now, shortest distance from origin (0,0) to tangent T will be ‌