Find the equation to the circle which touches each axis at a distance $5$ from the origin.

Open in App

Solution

Radius $= 5,$ Center $= (5, 5)$
$I s t$ Quadrant -
$\Rightarrow {(x - 5)}^{2} + {(y - 5)}^{2} = 25$
$I I n d$ Quadrant -
$\Rightarrow {(x + 5)}^{2} + {(y - 5)}^{2} = 25$
$I I I r d$ Quadrant -
$\Rightarrow {(x + 5)}^{2} + {(y + 5)}^{2} = 25$
$I V t h$ Quadrant -
$\Rightarrow {(x - 5)}^{2} + {(y + 5)}^{2} = 25$
Hence, solved.

Suggest Corrections

Similar questions

The equation of the circle in the first quadrant which touches each axis at a distance 5 from the origin is

Q. Find the equation to the circle which touches the axis of

x

at a distance

3

from the origin and intercepts a distance

6

on the axis of

y

Q. Find the equation of a circle which touches the positive axis of

y

at a distance

3

from the origin and intercepts a distance

6

on the axis of

x .

Q. Find the equation of the circles which touches the x -axis at a distance of 3 from the origin and center at

(0, 4)

Find the equation of a circle.
(i) which touches both the axes at a distance of 6 units from the origin.
(ii) which touches x-axis at a distance 5 from the origin and radius 6 units.
(iii) which touches both the axes and passes through the point (2, 1).

(iv) passing through the origin, radius 17 and ordinate of the centre is -15.

Join BYJU'S Learning Program

Find the equation to the circle which touches each axis at a distance 5 from the origin.

Radius =5, Center =(5,5)Ist Quadrant -⇒(x−5)2+(y−5)2=25IInd Quadrant - ⇒(x+5)2+(y−5)2=25IIIrd Quadrant -⇒(x+5)2+(y+5)2=25IVth Quadrant -⇒(x−5)2+(y+5)2=25Hence, solved.

Find the equation to the circle which touches each axis at a distance $5$ from the origin.