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

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$.

(i) Let the required equation of the circle be ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$.
It is given that the circle passes through the points (6, 0) and (0, 6).

∴ ${\left(6-h\right)}^2+{\left(0-k\right)}^2=6^2$
And, ${\left(0-h\right)}^2+{\left(6-k\right)}^2=6^2$

$$\begin{gathered} \Rightarrow {\left(6-h\right)}^2+{\left(-k\right)}^2=36 \\ \Rightarrow 36+h^2-12h+k^2=36 \\ \Rightarrow h^2+k^2=12h...\left(1\right) \end{gathered}$$

Also, $h^2+36+k^2-12k=36$
$\Rightarrow h^2+k^2=12k$ ...(2)

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

$12k=12h\Rightarrow h=k$

∴ From equation (2), we have:
$$\begin{gathered} k^2+k^2=12k \\ \Rightarrow k^2-6k=0 \\ \Rightarrow k\left(k-6\right)=0 \\ \Rightarrow k=6\left(\because k>0\right) \end{gathered}$$

Consequently, we get:
h = 6

Hence, the required equation of the circle is ${\left(x-6\right)}^2+{\left(y-6\right)}^2=36$ or $x^2+y^2-12x-12y+36=0$.

(ii) Let the required equation of the circle be ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$ .
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 ${\left(x-5\right)}^2+{\left(y-0\right)}^2=6^2$ or $x^2+y^2-10x-11=0$.


(iii) Let the required equation of the circle be ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$.
It is given that the circle touches both the axes.

Thus, the required equation will be $x^2+y^2-2ax-2ay+a^2=0$.

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

∴$4+1-4a-2a+a^2=0$
$$\begin{gathered} \Rightarrow a^2-6a+5=0 \\ \Rightarrow a^2-5a-a+5=0 \\ \Rightarrow a=1,5 \end{gathered}$$

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

(iv) Let the required equation of the circle be ${\left(x-h\right)}^2+{\left(y-k\right)}^2=a^2$.
Given:
k = −15, a = 17

The circle passes through the point (0, 0).
∴ Equation of the circle:
${\left(0-h\right)}^2+{\left(0-15\right)}^2={\left(17\right)}^2$
⇒ $h=\pm 8$

Hence, the required equation of the circle is ${\left(x-8\right)}^2+{\left(y+15\right)}^2={17}^2$ or ${\left(x+8\right)}^2+{\left(y+15\right)}^2={17}^2$, i.e. $x^2+y^2\pm 16x+30y=0$.