# find the equation of the circle whose centre is (3,-1) and which cuts off an intercept of length 6 from the line 2x-5y+18=0

Bishal
20 Points
5 years ago
we know that, the equation of the circle in standard form is-
(x-h)^2+(y-k)^2=r^2
the problem provides the co-ordinate (h,k) of centre of the circle.Hence,we xhould substitute 3 for h and -1 for k in equation above such that
(x-3)^2+(y+1)^2=r^2
here the radius is not given but we may find it by using the information provided by the problem.
The line that has the equation,  2x-5y+18=0 and the length of 6 is represent as the chord of  circle.
now, we may draw a perpendicular from the centre of the circle to this chord.
We may find the length of the perpendicular using the formula,
d=Ia.x_c+b.y_c+c I/square root of a^2+b^2

Notice that ax+by+c represents the equation of the chord and x_c and x_c co-ordinates of the centre.
d=I 6+5+18 I/square root of 4+25
=I 29 I/square root of 29
=square root of 29
Notice that the radius of the circle represents the hypotenuse of right angle traingle that has the lengths of the legs square root of 29 and 3.
Now , using Pythagorean theorem ,
r^2= 29+9
= 38
Now, we write the equation of the circle such as-
(x-3)^2+(y+1)^2=38
Hence,evaluating tyhe equation of the circle such that,
(x-3)^2+(y+1)^2=38 ..