Circles Part 1

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
‌

|2.0−0±√30|

√22+12

=‌

√30

√5

=√6