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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.

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Solution

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be x-h2+y-k2=a2.

(i) Let the required equation of the circle be x-h2+y-k2=a2.
It is given that the circle passes through the points (6, 0) and (0, 6).

6-h2+0-k2=62
And, 0-h2+6-k2=62

6-h2+-k2=3636+h2-12h+k2=36h2+k2=12h ...(1)

Also, h2+36+k2-12k=36
h2+k2=12k ...(2)

From (1) and (2), we get:

12k=12hh=k

∴ From equation (2), we have:
k2+k2=12kk2-6k=0kk-6=0k=6 k>0

Consequently, we get:
h = 6

Hence, the required equation of the circle is x-62+y-62=36 or x2+y2-12x-12y+36=0.

(ii) Let the required equation of the circle be x-h2+y-k2=a2 .
It is given that the circle with radius 6 units touches the x-axis at a distance of 5 units from the origin.
∴ a = 6, h = 5

Hence, the required equation is x-52+y-02=62 or x2+y2-10x-11=0.


(iii) Let the required equation of the circle be x-h2+y-k2=a2.
It is given that the circle touches both the axes.

Thus, the required equation will be x2+y2-2ax-2ay+a2=0.

Also, the circle passes through the point (2, 1).

4+1-4a-2a+a2=0
a2-6a+5=0a2-5a-a+5=0a=1,5

Hence, the required equation is x2+y2-2x-2y+1=0 or x2+y2-10x-10y+25=0.

(iv) Let the required equation of the circle be x-h2+y-k2=a2.
Given:
k = −15, a = 17

The circle passes through the point (0, 0).
∴ Equation of the circle:
0-h2+0-152=172
h=±8

Hence, the required equation of the circle is x-82+y+152=172 or x+82+y+152=172, i.e. x2+y2±16x+30y=0.

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