# Find the unique point that is the center of a circle passing through (2,5) and touching the line y = -7

*print*Print*list*Cite

Expert Answers

justaguide | Certified Educator

The locus of the center of a circle has to be found that passes through (2, 5) and touches the line y = -7. The line y = -7 is a tangent of the circle.

The general equation of a circle is x^2 + y^2 + ax + by + c = 0

As the circle passes through (2, 5)

=> 4 + 25 + 2a + 5b + c = 0

=> 2a + 5b + c + 29 = 0 ...(1)

At y = -7, the equation has only one root

x^2 + 49 + ax - 7b + c = 0 has only one root, this gives:

a^2 = 4*1*(c + 49 - 7b) ...(2)

From (1) and (2) it can be seen that there are 3 variables but only 2 equations.

This allows for an infinite number of solutions for a, b and c.

**The center of the circle satisfying the given conditions is not a unique point.**