Q. 144.2( 71 Votes )

# Find the equation of a circle with centre (2,2) and passes through the point (4,5).

Answer :

The centre of the circle is given as (h,k) = (2,2).

Since, the circle passes through point (4,5), the radius (r) of the circle is the distance between the points (2,2) and (4,5).

r =

=

=

=

Thus, the equation of the circle is

(x– h)^{2}+ (y – k)^{2} =r^{2}

(x –h)^{2} + (y – k)^{2} =

(x –2)^{2} + (y – 2)^{2} =

x^{2} - 4x+4 +y^{2}- 4y_{+}4 =13

** x ^{2} +y^{2} – 4x -4y = 5**

Rate this question :

Show that the quadrilateral formed by the straight lines x – y = 0, 3x + 2y = 5, x – y = 10 and 2x + 3y = 0 is cyclic and hence find the equation of the circle.

RS Aggarwal - MathematicsIf ( - 1, 3) and (∝, β) are the extremities of the diameter of the circle x^{2} + y^{2} – 6x + 5y – 7 = 0, find the coordinates (∝, β).

Find the equation of the circle which passes through the points A(1, 1) and B(2, 2) and whose radius is 1. Show that there are two such circles.

RS Aggarwal - MathematicsFind the equation of the circle concentric with the circle x^{2} + y^{2} – 6x + 12y + 15 = 0 and of double its area.