Question

Find the equations of circle passing through the points (1,1),(2,2) and whose radius is 1 units?​

Answer 1

equation of circle :-

(x-1)^2 + (y-2)^2 = 1

or

(x-2)^2 + (y-1)^2 = 1

Answer 2

The equation of circle would be $(x-2)^2+(y-1)^2=1$ or $(x-1)^2+(y-2)^2=1$

Step-by-step explanation:

Consider the equation of the circle is,

$(x-h)^2+(y-k)^2=r^2$

where, (h,k) is the center of the circle and r is the radius.

Here, r = 1,

So, the equation of the circle is,

$(x-h)^2+(y-k)^2=1$

If the circle passes through the point (1,1), then equation of the circle satisfied by the point.

That is, $(1-h)^2+(1-k)^2=1$              ...... (1)

Similarly, circle passes through the point (2, 2),

So, $(2-h)^2+(2-k)^2=1$                  ...... (2)

From equations (1) and (2),

$(1-h)^2+(1-k)^2=(2-h)^2+(2-k)^2$

$1+h^2-2h+1+k^2-2k=4-4h+h^2+4-4k+k^2$

$2-2h-2k=8-4h-4k$

$-2h-2k=8-4h-4k-2$

$-2h=6-4h-4k+2k$

$-2h+4h=6-2k$

$2h=6-2k$

$h=3-k$                                           ...... (3)

From equation (1) and (3),

$(1-3+k)^2+(1-k)^2=1$

$(k-2)^2+(1-k)^2=1$

$k^2+4-4k+1+k^2-2k=1$

$2k^2-6k+4=0$

$k^2-3k+2=0$

$k^2-2k-k+2=0$

$k(k-2)-1(k-2)=0$

$(k-1)(k-2)=0$

By zero product property,

k = 1 or k = 2

If k = 1, then from equation (3),

h=3-1=2

Thus, equation of the circle is,

$(x-2)^2+(y-1)^2=1$

If k = 2, then from equation (3),

h=2-1=1

Thus, equation of the circle is,

$(x-1)^2+(y-2)^2=1$

#Learn more:

The points (7,3), (2,8), (-3,3) lie on a circle. Find the  equation of the circle and its radius.

https://brainly.in/question/15148745