Question

For what values of k the points (0,0),(1,3),(2,4) and (k,3) are concyclic

Answer 1

Answer:

k=9

Step-by-step explanation:

We are given that (0,0),(1,3),(2,4) and (k,3) are concyclic.

We have to find the value of k.

When the points are concyclic it means the circle.

Equation of circle is given by

$x^2+y^2+2fx+2gy+c=0$

When (0,0) lie on the circle

Then,we get c=0

When (1,3) lies on the circles then we get

$1+9+2f+6g=0$

$2f+6g=-10$

$f+3g=-5$.......(1)

When the point (2,4) lies on the circle then we get

$4+16+4f+8g=0$

$4f+8g=-20$

$f+2g=-5$.....(2)

Subtract equation (1) from equation (2)

g=0

Substitute the value in equation (1)

$f+0=-5$

$f=-5$

When point (k,3) lies on the circle

$k^2+9+2kf+6g=0$

Substitute the values

$k^2+9-10k=0$

$k^2-10k+9=0$

$(k-1)(k-9)=0$

$k-1=0\implies k=1$

$k-9=0\implies k=9$

The point (1,3) or (9,3).

(1,3) is already given therefore, k=9