Question

Find the equation of the circle touching the lines x=0 y=0 and x=2c

Answer 1

Answer:
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${x}^{2} + {y}^{2} - 2xc - 2yc + {c}^{2} = 0$

Solution:
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Standard equation of circle having centre at (a,b) and radius r

${(x - a)}^{2} + ( {y - b)}^{2} = {r}^{2}$

Here the given problem is drawn in the attachment,here we can easily see that,centre of circle is (c,c) and it's radius is C.

here y= 0 is x -axis

x= 0 is y- axis

x= 2c,is a line parallel to y axis

So, equation of circle

$( {x - c)}^{2} + ( {y - c)}^{2} = {c}^{2} \\ \\ {x}^{2} - 2xc + {c}^{2} + {y}^{2} - 2yc + {c}^{2} = {c}^{2} \\ \\ {x}^{2} + {y}^{2} - 2xc - 2yc + 2 {c}^{2} = {c}^{2} \\ \\ {x}^{2} + {y}^{2} - 2xc - 2yc + {c}^{2} = 0$

is the equation of circle touching the lines x=0 y=0 and x=2c.

Hope it helps you.

Answer 2

Answer:

$(x-c)^2+(y-c)^2=c^2$

Step-by-step explanation:

It is given that the circle touching the lines x=0 y=0 and x=2c.

The circle lies between x=0 and x=2c it means the diameter of the circle is 2c.

Radius of the circle = c

x-coordinate of center = 0+c = c

y-coordinate of center = 0+c = c

Center of circle is (c,c).

The standard equation of the circle is

$(x-h)^2+(y-k)^2=r^2$

Where, (h,k) is center and r is radius.

The equation of circle is

$(x-c)^2+(y-c)^2=c^2$

Therefore, the equation of circle in standard form is $(x-c)^2+(y-c)^2=c^2$.