If the points P (1, 3) and Q (2, 6) lie on the boundary of a circle with the centre at O (5,y), then what is the value ofy?

can I use section formula for this question?

I tried to solve it using section formula but was unable to find the answer.

plsssssssssss.. clear my doubt as soon as possible

Answer :

As given points P ( 1 , 3 )  and Q ( 2 , 6 )  lies on the boundary of circle .
And
Center at O ( 5 , y ) , So
Here OP =  OQ  as these are the radius of this circle . So we use distance formula to get value of y , As :

We know distance formula d  = $\sqrt{{\left( x_2 - x_{1 }\right)}^2 + {\left( y_2 - y_{1 }\right)}^2}$
So,
OP  = $\sqrt{{\left(1 - 5 \right)}^2 + {\left( 3 - y\right)}^2}$
And
OQ = $\sqrt{{\left(2 - 5 \right)}^2 + {\left( 6 - y\right)}^2}$
As we know OP and OQ are radius of this circle , So

$\Rightarrow$$\sqrt{{\left(1 - 5 \right)}^2 + {\left( 3 - y\right)}^2}$  = $\sqrt{{\left(2 - 5 \right)}^2 + {\left( 6 - y\right)}^2}$

$\Rightarrow$$\sqrt{{\left(- 4 \right)}^2 + {\left( 3 - y\right)}^2}$  = $\sqrt{{\left( - 3 \right)}^2 + {\left( 6 - y\right)}^2}$

$\Rightarrow$$\sqrt{16 +9 + y^2 - 6y}$  = $\sqrt{9+ 36 + y^2 - 12y}$

Now we taking whole square on both hand side , and get

$\Rightarrow$16 + 9 + y² - 6y  = 9 + 36 + y² - 12y 

$\Rightarrow$6y  = 20

$\Rightarrow$y  = $\frac{20}{6}$

$\Rightarrow$y  = $\frac{10}{3}$  ( Ans )