144k views
3 votes
the line with equation y=kx intersects the circle with equation x^2-10x+y^2-12y+57=0 at two distinct points

1 Answer

4 votes

For the line y=kx to intersect the circle
x^2-10x+y^2-12y+57=0at two distinct points, the discriminant of the quadratic equation obtained by substituting y=kx into the circle's equation must be positive. Solving this inequality yields the range of possible values for k: 0.71<k<2.15.

Consider the line y=kx and the circle
x^2-10x+y^2-12y+57=0. For the line to intersect the circle at two distinct points, the quadratic equation obtained by substituting y=kx into the circle's equation should have two real and distinct roots.

This quadratic equation is
(1+k^2)x^2-(10+12k)x+57=0. The discriminant of this quadratic is responsible for determining the nature of its roots.

For two distinct roots, the discriminant must be greater than zero. This leads to the inequality
21k^2-60k+32 < 0, which can be solved to obtain the range of possible values for k: 0.71<k<2.15.

Complete question below:

The line with equation y=kx intersects the circle with equation
x^2-10x+y^2-12y+57=0 at two distinct points. Determine the range of possible values for k.

User Callyalater
by
5.8k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.