Question

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.

Answer 1

The circle touches the axes at (0, 6) and (6, 0) respectively
Thus, the centre of circle will be (6,6) and
radius =$OA=\sqrt{(6-0{)}^2+(6-6{)}^2}=\sqrt{36}=6$
(by distance formula)
$\therefore$ the equation of circle will be $(x-6{)}^2+(y-6{)}^2=6^2$
$\Rightarrow x^2+y^2-12x-12y+36=0$

(i) The circle touches the x - axis at A = (5, 0) and has radius 6 unit
Thus,
centre = (5, b)
By distance formula OA = 6
$\Rightarrow \sqrt{(5-5{)}^2+(b-0{)}^2}=6\Rightarrow b=6\Rightarrow centre=(5,6)$
So, the equation of required circle is
$(x-5{)}^2+(y-6{)}^2=6^2\Rightarrow x^2+y^2-10x-12y+25=0$

(iii) The circle touches both thb axis at A =(a, 0) and B (0, a)
so, the centre of circle will be (a, a) and radius= a
so, the equation of circle is $(x-a{)}^2+(y-a{)}^2{a}^2\dots \left(A\right)$
Now,
(A) Passes through P(2, 1)
$\therefore (2-a{)}^2+(1-a{)}^2=a^2\Rightarrow 4-4a+a^2+1-2a+a^2=a^2\Rightarrow 5-6a+a^2=0\Rightarrow a=5or1$

Thus the equation of circle will be
$x^2+y^2-10x-lOy+25=0or{x}^2+y^2-2x-2y+1=0$

(iv) The circle passes through origin (0, 0) and has radius = 17 units Also, die ordinate of centre is -15 then assume abssisa is a
$\therefore OC=17\Rightarrow \sqrt{(a-0{)}^2+(0-15{)}^2}=17$
(By distance formula)
$\Rightarrow \sqrt{a^2+225}=17\Rightarrow a^2+225=289\Rightarrow a^2=64\Rightarrow a=\pm 8\therefore centre=(\pm 8,-15)$
Thus the equation of circle will be,
$(x\pm 8{)}^2+(y+15{)}^2={17}^2\Rightarrow x^2+y^2\pm 16x+30y=0$