Let the lengths of intercepts on x-axis and y-axis made by the circle
x² + y² + ax + 2ay + c = 0, (a < 0) be 2√2 and 2√5 , respectively. Then the shortest distance from origin to a tangent to this circle which is perpendicular to the line x + 2y = 0, is equal to :

Solution

Maths sir · 5 steps

Use the intercepts to find the circle, then notice the origin sits on the centre's line of slope 2.

1. Intercept formulas. For x²+y²+2gx+2fy+c=0, the x-intercept is 2√(g²-c) and the y-intercept is 2√(f²-c). Here g=a/2 and f=a, so a²/4-c=2 and a²-c=5.
2. Find the circle. Subtracting gives 3a²/4=3, so a²=4 and a=-2 since a<0. Then c=1-2=-1. The circle is x²+y²-2x-4y-1=0, with centre (1,2) and radius √(1+4+1)=√6.
3. Tangent lines. The line x+2y=0 has slope -1/2, so the tangent has slope 2. Write it as 2x-y+k=0 and set its distance from the centre equal to the radius.

   (|2(1)-2+k|)/√5=√6 ⇒ |k|=√30

4. Distance from origin. The distance from (0,0) to 2x-y±√30=0 is √30/√5=√6 for both tangents. This happens because the line 2x-y=0 passes through both the origin and the centre (1,2).
5. So. The shortest distance from the origin to such a tangent is √6, option (b).

The idea to keep. Intercept lengths are 2√(g²-c) and 2√(f²-c). For a tangent of known slope, write y=mx+k and match the distance from the centre to the radius.

Answer from the NTA final answer key. Solution by PrepWiser.