Question

9. Find the equation of a circle touching the X-axis and equation diameters are x - y = 1 1and 2x + y = 5 .​

Answer 1

Answer:

x - y = 1

2x + y = 5

+

3x + 0 = 6

3x = 6

x = 6/3

x = 2

therefore, 2x + y = 5

2×2 + y = 5

4 + y = 5

y = 5 - 4

y = 1

Answer 2

Appropriate Question :-

Find the equation of a circle touching the X-axis and equation diameters are x - y = 1 and 2x + y = 5.

$\large\underline{\sf{Solution-}}$

Given that equation of diameters of circle are

$\rm \: x - y = 1 - - - (1)$

and

$\rm \: 2x + y = 5 - - - (2)$

We know, the diameters of circle intersects at center of circle. So, point of intersection of diameters give the coordinates of center of circle.

So, on adding equation (1) and (2), we get

$\rm \: 3x = 6$

$\rm\implies \:x = 2$

On substituting x = 2, in equation (1), we get

$\rm \: 2 - y = 1$

$\rm \: - y = 1 - 2$

$\rm \: - y = - 1$

$\rm\implies \:y = 1$

$\rm\implies \:Coordinates \: of \: center, \: O = (2,1)$

Now, further given that circle touches the x - axis. So, as we know that radius and x - axis are perpendicular to each other. So, radius of circle, r = Distance of center from x - axis = 1 unit.

So, required equation of circle having center (2, 1) and radius, r = 1 is given by

$\rm \: {(x - 2)}^{2} + {(y - 1)}^{2} = {1}^{2}$

$\rm \: {x}^{2} + 4 - 4x + {y}^{2} + 1 - 2y = 1$

$\rm \: {x}^{2} + 4 - 4x + {y}^{2} - 2y = 0$

$\rm \: {x}^{2} + {y}^{2} - 4x - 2y + 4= 0$

Hence, required equation of circle is

$\color{green}\rm\implies \:\boxed{ \rm{ \:\rm \: {x}^{2} + {y}^{2} - 4x - 2y + 4= 0 \: \: }}$

$\rule{190pt}{2pt}$

Formula Used :-

Equation of circle having center (h, k) and radius r is given by

$\color{green}\boxed{ \rm{ \: {(x - h)}^{2} + {(y - k)}^{2} = {r}^{2} \: \: }}$

Remark :- Green and purple lines shows equation of first and second diameter.