The line of shortest distance must be normal to both curves. If P is the optimal point on the parabola and C is the centre of circle then CP is || to parabola normal at P.
Circle is (x+3)²+(y−6)²=25 so C is (−3,6)
Let P be (p²,2p) so tangent direction is 1/p and normal direction is −p
∴ (2p−6) / (p²+3) = −p → p³+5p−6=0 → (p−1)(p²+p+6) = 0
This has only one real root p=1 which gives P as (1,2)
The min distance is one of CP±5 = √{ 1+3)²+(2−6)² } ± 5 = 4√2±5
Min distance is 4√2−5
Circle is (x+3)²+(y−6)²=25 so C is (−3,6)
Let P be (p²,2p) so tangent direction is 1/p and normal direction is −p
∴ (2p−6) / (p²+3) = −p → p³+5p−6=0 → (p−1)(p²+p+6) = 0
This has only one real root p=1 which gives P as (1,2)
The min distance is one of CP±5 = √{ 1+3)²+(2−6)² } ± 5 = 4√2±5
Min distance is 4√2−5
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this is interesting, someone shd help out...