The side PQ of triangle ΔPQS is a tangent of the circle C₂, given that
point Q is on the surface of circle C₂.
Correct response:
- QS is not perpendicular to tangent PQ, therefore, QS is not a diameter of circle C₂
Methods used to prove the property of QS
The given parameters are;
Points on the circle are; Q, P, and S
Tangent to circle, C₁ = RST
Point through which circle C₂ passes = Center O
Required:
To prove that SQ is not a diameter of circle C₂
Solution:
Given that RST is a tangent, we have;
OS is perpendicular to RST by definition of a tangent to a circle
Therefore;
90° = ∠OSQ + ∠QST
Which gives;
∠OSQ = 90° - 46° = 44°
ΔQOS is an isosceles triangle, by definition of isosceles triangles
Therefore;
∠QOS = 180° - 2 × 44° = 92°
∠QPS = 0.5 × 92° = 46° Angle at center is twice angle at the circumference
∠SOP = [tex]\mathbf{\frac{1}{2}}[/tex] × ∠SQP
Let x represent the base angles of ΔSOP, we have;
∠SOP = 180° - 2·x
Therefore;
[tex]\angle SQP = \dfrac{180^{\circ} - 2\cdot x}{2} = \mathbf{90^{\circ} - x}[/tex]
Which gives;
∠SQP = 90° - x < 90°
PQ is a tangent to circle, C₂, by definition of a tangent (number of points circle C₂ intersect PQ is one)
PQ is not perpendicular to QS, which gives;
- QS is not made up of two radii, and therefore, QS is not a tangent of circle, C₂
Learn more about a tangent of a circle here:
https://brainly.com/question/26080631
