Hi !!

Please tell me how to solve the following question :

The length of the diameter of the circle which touches the X axis at the point (1,0) and passes through the point (2,3) is ?

If a circle touches the x-axis, its equation is

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

but it is given that the equation passes thru the point (1,0)

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

=> h=1,0

and also that tha equation passes thru the point (2,3)

=> 4+9-4h-6k+h^2=0

=> 13-4-6k+1=0

=> k=5/3=radius

diameter = 2(radius)= 10/3

let (2,3) be A, and (1,0) be B and the centre of the circle be C,then

since the cirlce touches the x axis at B.hence the x axis is a tangent to the cirle,

and the C lies on the line perpendicular to x axis .i.e. x=1

now the co ordinates of C are (1,y).

applying distance formula for the centre from two points on circumference,we get

(AC)^2=(BC)^2

solving we get y=5/3

hence C=(1,5/3)

radius = BC=5/3

diameter=10/3 (ANS)

hope i hepled you.

Let the C(h,k)

We know that for a circle touching x-axis, modulus of y-co-ordinate is equal to the radius.

=>(h-1)² + k² = k²

=>h=1

Again: 

(h-2)²   + (k-3)² = k²

h=1  =>k=5/3.

=>Diameter=10/3 units.

the ans is 10/3

10/3