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

Solution :

Let us assume that the coordinates of the center of the circle are C(h,k) and its radius is r.

Now, since the circle touches X-axis at (1,0), hence its radius should be equal to ordinate of center.

\(\implies\) r = k

Hence, the equation of circle is \((x – h)^2 + (y – k)^2\) = \(k^2\)

Also, given that the circle passes through points (1, 0) and (2, 3). Hence, substituting them, in the equation of circle we get

\((1 – h)^2 + (0 – k)^2\) = \(k^2\)       ……(i)

\((2 – h)^2 + (3 – k)^2\) = \(k^2\)        ……(ii)

from equations (i) and (ii), we get

k = \(5\over 3\)

Hence, The diameter of the circle is 2k = \(10\over 3\)


Similar Questions

The equation of the circle passing through the foci of the ellipse \(x^2\over 16\) + \(y^2\over 9\) = 1 and having center at (0, 3) is

The circle passing through (1,-2) and touching the axis of x at (3, 0) also passes through the point ?

Let C be the circle with center at (1,1) and radius 1. If T is the circle centered at (0,y) passing through origin and touching the circle C externally, then the radius of T is equal to

The equation of the circle through the points of intersection of \(x^2 + y^2 – 1\) = 0, \(x^2 + y^2 – 2x – 4y + 1\) = 0 and touching the line x + 2y = 0, is

Find the equation of circle having the lines \(x^2\) + 2xy + 3x + 6y = 0 as its normal and having size just sufficient to contain the circle x(x – 4) + y(y – 3) = 0.

Leave a Comment

Your email address will not be published.