Question

Find the equation of the circle which passes through the points (1,3) and (2,-1), and has its centre on the line 2x+y-4=0

Answer 1

Answer

Step-by-step explanation:

Answer 2

Given,

$x_{1}$ = 1

$x_{2}$ = 2

$y_{1}$ = 3

$y_{2}$ = -1

To find,

The equation of the circle with center on the line $2x+y-4=0$

Solution,

In order to solve the given question correctly, we can follow the given steps.

From our knowledge, we have of this chapter on Mathematic we know that,

The formula for the equation of a circle is,

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

Here,

h and k are the centers of the circle

r is the radius of the circle

Next, we have to substitute points (1, 3) and (2, - 1) in the formula given above likewise,

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

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

As both these equations are equal $r^{2}$ we can equate them,

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

If we open the brackets and solve according to the formula we have,

∴ $(a+b)^{2}=a^{2}+2ab+b^{2}$

$1-2h+h^{2}+9-6k+k^{2} = 4-4h+h^{2}+1+2k+k^{2}$

$-2h-6k+10 = -4h+2k+5$

$2h=8k-5$

Substitute x = h and y = k in 2x+y-4 = 0 as centre lies on this line,

$2h=4-k$

Now we can easily equate 2h

$8k-5=4-k$

$9k=9$

$k=1$

If we substitute $k=1$ in $2h=4-k$

$2h=4-k$

$2h=3$

$h=1.5$

Now, substitute $k=1$ and $h=1.5$ in the equation $(1-h)^{2}$ + $(3-k)^{2}$ = $r^{2}$  

$(1-1.5)^{2}+(3-1)^{2}=r^{2}$

$0.5^{2} +2^{2} =r^{2}$

$0.25+4=r^{2}$

$4.25=r^{2}$

Now finally, we can substitute $k=1$, $h=1.5$ and $r^{2}=4.25$ in the equation

$(x-h)^{2}$ + $(y-k)^{2}$ = $r^{2}$ to obtain the equation of the circle

$(x-1.5)^{2}+(y-1)^{2}=4.25$

Hence, the equation of the given circle is $(x-1.5)^{2}+(y-1)^{2}=4.25$