# How do you find the equation of a circle that touch the x axis and passes through (1,1) and has its centre on the line x + y = 3?

Aug 21, 2016

There are $2$ circles, ${S}_{1} , \mathmr{and} , {S}_{2}$, satisfying the given conds.

${S}_{1} : {x}^{2} + {y}^{2} + 4 x - 10 y + 4 = 0$,

and,

 S_2 : x^2+y^2-4x-2y+4+0".

#### Explanation:

Let $C \left(h , k\right)$ be the centre, and, $r$ the radius of the reqd. circle $S$.

The given pt.$P \left(1 , 1\right) \in S \Rightarrow C {P}^{2} = {r}^{2} \Rightarrow {\left(h - 1\right)}^{2} + {\left(k - 1\right)}^{2} = {r}^{2.} . . \left(1\right)$

$C \in \left[x + y = 3\right] \Rightarrow h + k = 3. \ldots \ldots \ldots \left(2\right)$

Finally, $S$ touches $\text{X-axis : y=0}$

":. the bot-distance from $C \left(h , k\right)$ to the $\text{X-axis} = r$.

$\therefore | k | = r \ldots \ldots \ldots \ldots \ldots \left(3\right)$

From $\text{(1) & (3)} , {\left(h - 1\right)}^{2} + {\left(k - 1\right)}^{2} = {k}^{2}$.

BY $\left(2\right)$, then, ${\left(3 - k - 1\right)}^{2} + {\left(k - 1\right)}^{2} - {k}^{2} = 0$, or,

${\left(2 - k\right)}^{2} + {\left(k - 1\right)}^{2} - {k}^{2} = 0$

${k}^{2} - 6 k + 5 = 0 \Rightarrow \left(k - 5\right) \left(k - 1\right) = 0 \Rightarrow k = 5 , \mathmr{and} , 1$.

Accordingly, $\text{h=-2, or, 2, and, finally, r=5, or, 1}$.

Thus, we have $2$ circles ${S}_{1} , \mathmr{and} , {S}_{2}$:

${S}_{1} : C e n t r e {C}_{1} \left(- 2 , 5\right) , r = 5$, with the eqn. given by,

${S}_{1} : {\left(x + 2\right)}^{2} + {\left(y - 5\right)}^{2} = {5}^{2} , i . e . , {x}^{2} + {y}^{2} + 4 x - 10 y + 4 = 0$,

and,

S_2 : Centre C_2(2,1), r=1, eqn. x^2+y^2-4x-2y+4+0".

Enjoy Maths.!