Question

Find the equation of a circle passing through point (4,6) whose diameter are along x+2y-5=0 and 3x-y-1 =0 ?

Answer 1

Answer:

Equation of the circle is $(x-1)^2+(y-2)^2=25$

Step-by-step explanation:

Given equations of lines are:

$x+2y-5=0$

or, $x+2y=5$         ................... (1)

And $3x-y-1=0$

or, $3x-y=1$           ................... (2)

Since the diameters lie along these lines

Therefore, the intersection of these lines will be the centre of the circle

To find the intersection of these lines we simply need to solve these two equations as simultaneous linear equations

Multiplying eq (2) by 2 and adding it to the eq (1)

$7x=7$

or, $x=1$

Putting the value of x in eq (2)

$3-y=1$

$\implies y=2$

Thus the centre of the circle is $(1, 2)$

Since the circle passes through the point $(4,6)$

Therefore, the distance between the centre and this point will be the radius of the circle

Therefore, radius r

$=\sqrt{(4-1)^2+(6-2)^2}$

$=\sqrt{9+16}$

$=\sqrt{25}$

$=5$

Equation of the circle if its centre is (h,k) and radius r is given by

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

Thus, the equation of the circle is

$(x-1)^2+(y-2)^2=5^2$

or, $(x-1)^2+(y-2)^2=25$

Hope this helps.