Question

Find the equation of the circle which touches the axis of:
(a) $x$ at a distance $+3$ from the origin and intercepts a distance $6$ on the axis of $y$.
(b) $y$ at a distance $-3$ from the origin and intercepts a length $8$ on the axis of $x$.
(c) $x,$ pass through the point $(1,1)$ and have linie $x+y=3$ as diameter.

Answer 1

(a)

Let $O$ be the centre of axis of circle & $\times$ bisect the intercept $AB\left(6units\right)$ i.e $XA=3units$

In $△XOA,O{X}^2+X{A}^2=O{A}^2$ [Pythagoras theorem]

$\Rightarrow 9+9=O{A}^2$

$OA=$ radius $=\sqrt{18}$ ....... $\left(1\right)$

Centre of circle $+(3,3\sqrt{2})$

Equation: $(x-3{)}^2+(y-3\sqrt{2}{)}^2=18$

(b)

Similarly in $△XOB$

$O{X}^2+X{B}^2=O{B}^2$ [Pythagoras]

$9+16=O{B}^2$

$OB=r=5.....\left(2\right)$

Center of circle $=(-5,-3)$

Equation: $(x+5{)}^2+(y+3{)}^2=25$

(c)

Let $O$ be the centre of circle with coordinates $(h,k)$. The circle then touches $x-$ axis at $(h,o)$ Thus, radius $r=k$

As, $(h,k)$ be $x+y=3$

$\Rightarrow K=3-h$

Now, $A{O}^2=r^2$

$(h-1{)}^2+(3-h-1{)}^2=(3-h{)}^2$

$\Rightarrow h^2+1-2k+4+h^2-4k=9+h^2-6h$

$h^2=4\Rightarrow h=\pm 2$

$K=3-h=5,1$

Thus centre can be $(-2,5)$ or $(2,1)$

Required equation: $\begin{array}{c}(x+2{)}^2+(y-5{)}^2=25\\ (x-2{)}^2+(y-1{)}^2=1\end{array}\begin{array}{c}[r=k=5]\\ [r=k=1]\end{array}$