1. Consider a circle whose center is on the \( x \)-axis. If a line given by \( y=x \) is tangent to the circle at a point \( (2,2) \), then what is the equation of the circle? (a) \( x^{2}+y^{2}=8 \) (b) \( (x-2)^{2}+y^{2}=4 \) (c) \( (x-4)^{2}+y^{2}=8 \) (d) \( (x-1)^{2}+y^{2}=5 \)
Added by Nicol-S M.
Close
Step 1
Let's denote the center as (h, 0). Second, we know that the line y = x is tangent to the circle at the point (2,2). This means that the radius of the circle is the distance from the center of the circle to the point of tangency. The distance between two points Show more…
Show all steps
Your feedback will help us improve your experience
Anurag Kumar and 56 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
A tangent to the circle $x^{2}+y^{2}=1$ through the point $(0,5)$ cuts the circle $x^{2}+y^{2}=4$ at $A$ and $B$. The tangents for the circle $x^{2}+y^{2}=4$ at $A$ and $B$ meet at $C$. The coordinates of $C$ are (A) $\left(\frac{8 \sqrt{6}}{5}, \frac{4}{5}\right)$ (B) $\left(-\frac{8 \sqrt{6}}{5}, \frac{4}{5}\right)$ (C) $\left(\frac{8 \sqrt{6}}{5},-\frac{4}{5}\right)$ (D) $\left(-\frac{8 \sqrt{6}}{5},-\frac{4}{5}\right)$
Which of the following is the equation of the circle that has its center at the origin and is tangent to the line with equation $3 x-4 y=10 ?$ (A) $x^{2}+y^{2}=2$ (B) $x^{2}+y^{2}=4$ (C) $x^{2}+y^{2}=3$ (D) $x^{2}+y^{2}=5$ (E) $x^{2}+y^{2}=10$
The length of the tangent from any point on the circle $x^{2}+y^{2}-8 x+5 y-4=0$ to the circle $x^{2}+y^{2}-8 x+5 y=0$ is (a) 1 (b) 2 (c) 4 (d) 6
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Watch the video solution with this free unlock.
EMAIL
PASSWORD