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 circle's equations $x^{2} + y^{2} - 10 x - 10 y + 41 = 0, x^{2} + y^{2} - 24 x - 10 y + 160 = 0$

Open in App

Solution

Given :
$S_{1} : x^{2} + y^{2} - 10 x - 10 y + 41 = 0 S_{2} : x^{2} + y^{2} - 24 x - 10 y + 160 = 0$
Centre and radius of the circles are
$C_{1} = (5, 5), r_{1} = 3 C_{2} = (12, 5), r_{2} = 3 C_{1} C_{2} = 7$

So minimum distance between $P_{1}$ and $P_{2}$
$= C_{1} C_{2} - (r_{1} + r_{2}) = 7 - (3 + 3) = 1$

Suggest Corrections

Similar questions

P_{1}

P_{2}

P_{1}

P_{2}

x^{2} + y^{2} - 10 x - 10 y + 41 = 0, x^{2} + y^{2} - 24 x - 10 y + 160 = 0

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

x^{2} + y^{2} - 22 x - 10 y + 137 = 0

x^{2} + y^{2} + 2 g_{1} x - a^{2} = 0

x^{2} + y^{2} + 2 g_{2} x - a^{2} = 0

p_{1}, p_{2}

(0, a)

(0, - a)

p_{1}, p_{2} =

5

y = x

Join BYJU'S Learning Program