touches both the axes lying in 4th quadrant and has it center on the line x-2y=3
Answers
Answer:
Circle touches X & Y axis ( given)
=> Its centre should be equidistant from both axis. ( & the distance = radius of the circle)
=> modulus of both coordinates of its centre will be equal.
Also centre of the circle lies on the equation x-2y = 3
=> coordinates of the centre satisfy the above equation.
Since circle can be in any quadrants, =>coordinates of the centre will be either (a,a) , (-a,-a) , (a,-a) or(-a,a)
Equation is x-2y = 3
=> a -2a= 3 ( by taking centre (a,a)
=> -a = 3
=> a = -3
=> centre of the circle is (-3,-3) , so radius = 3 unit . . . . . . . . . (1)
Now, by taking coordinates of the centre as
(a,-a)
We get x-2y = 3
=> a - 2* -a = 3
=> a + 2a = 3
=> 3a = 3 => a = 1
So, coordinates of the centre = ( 1, -1), & radius = 1 unit . . . . . . . . . . . . . .(2)
With other centres ( -a,-a) & (-a,a) we get the same value.
Now, apply the standard form of the equation of a circle = ( x- h)² + ( y-k)² = r² , where h & k are the coordinates of the centre & r is radius
=> equation of circle in the 3rd quadrant =
(x+3)² + (y+3)² = 3²
=> x² +6x+9 + y²+6y + 9 = 9
=> x² + y² + 6x + 6y +9 = 0
& equation of circle in the 4th quadrant=
( x-1)² + (y+1)² = 1²
=> x² -2x +1 + y² +2y + 1 = 1
=> x² + y² -2x + 2y +1 = 0
Step-by-step explanation:
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