Question

A circle having its center in the first quadrant touches y axis at point ( 0 2 ) and passes through point (1 0 )

Answer 1

Answer:

equation of circle is given by when h and k are the centre of the circle and r is the radius

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

given circle touches the y-axis

so h=r

so (x-h)^2+(y-k)^2=h^2

and the circle passing through (1,0) and(0,2)

so for (1,0) ,

(1-h)^2+(0-k)^2=h^2

=>1+h^2-2h+k^2=h^2

=>1-2h+k^2=0

for (0,2)

(0-h)^2+(2-k)^2=h^2

=>h^2+4-4k+k^2=h^2

=>4-4k+k^2=0

=>(k-2)^2=0

=>k-2=0

=>k=2

putting the value of k in 1-2h+k^2=0 we get

1-2h+4=0 => 2h=5 => h=5/2

hence putting the value of h and k in the standard equation we get

(x-5/2)^2+(y-2)^2=(5/2)^2

=>(2x-5)^2+(y-2)^2=25

=>(4x^2+25-20x+y^2+4-4y)=25

=>4x^2+y^2-20x-4y+4=0