JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Circle

The minimum distance between any two points

P_{1}

and

P_{2}

while considering point

P_{1}

on one circle and point

P_{2}

on the other circle for the given circles' equations

x^{2} + y^{2} - 10 x - 10 y + 41 = 0

x^{2} + y^{2} - 24 x - 10 y + 160 = 0

is ____

Answer: 1

Solution

Completing the square,

x^{2} + y^{2} - 10 x - 10 y + 41 = 0

(x - 5)^{2} + (y - 5)^{2} = 9

, a circle of centre

(5, 5)

and radius

3

, and

x^{2} + y^{2} - 24 x - 10 y + 160 = 0

(x - 12)^{2} + (y - 5)^{2} = 9

, a circle of centre

(12, 5)

and radius

3

. The distance between the centres is

7

, which exceeds

3 + 3 = 6

, so each circle lies entirely outside the other and the shortest distance between a point of one and a point of the other is measured along the line of centres:

7 - 3 - 3 = 1.

The minimum distance is

1

JEE Main · 2021 · 17 Mar 2021 · Shift 1 · Integer answer

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