Question

find the coordinates  of the points which is at a distance of 2 unitsfrom (5,4)and 10 units from (11,-2).

Answer 1

Here is ur ans.
let the coordinates of the point A(x,y)

B:(5,4) and C :(11,-2)

AB = 2 units

AB2 = 4

therefore by distance formula

(5 - x)2 + (4 - Y)2 = 4

25 + x2 - 10x + 16 + y2 - 8y = 4

x2 + y2 - 10x - 8y = -37

x2 + y2 = 10x + 8y -37 .......(i)



also AC2 = 100

therefore by distance formula

( 11 - x)2 + ( 2 + y)2 = 100

121 + x2 - 22x + 4 + y2 + 4y = 100

x2 + y2 - 22x + 4y = - 25

x2 + y2 = 22x - 4y - 25 .......(ii)

from (i) and (ii)

22x - 4y - 25 = 10x + 8y - 37

12x - 12y + 12 = 0

x - y = -1

x = y - 1....(iii)

putting (iii) in (i) we get

( y -1)2 + y2 = 10(y - 1) + 8y - 37

2y2 -2y + 1 = 10y - 10 + 8y - 37

2y2 - 20y + 48 = 0

y2 - 10y + 24 = 0

(y - 6)(y - 4) = 0

y = 6 and y = 4

so x = 5 and x = 3

so coordinates of A is either (5,6) or (3,4)
Hope it helps u!!

Answer 2

The required coordinates are (3,4) and (5,6).

Step 1: Find the coordinates.

Given- point A(5,4) and B(11,-2)

Let P (x,y) be a point which is at a distance of 2 units from A and 10 units from B,

Using distance formula,

AP =2 units

$\sqrt{(x-5)^2 + (y-4)^2} = 2$

$(x-5)^2 + (y-4)^2 = 4$

$x^{2} +y^2 -10x-8y+37=0$                                    ..(i)

Also ,we have BP= 10 units

$\sqrt{(x-11)^2 + (y+2)^2} = 10$

$(x-11)^2 + (y+2)^2 =100$

$x^{2} +y^2-22x+4y+25=0$                                    .. (ii)

On solving equation (i) and (ii), we get

$x-y+1=0$

$y=x+1$

Using the values of y in the above equations, we get

$x=3,5$

$y=4,6$

Hence, the required coordinate is (3,4) and (5,6).