![McDougal Littell Jurgensen Geometry: Student Edition Geometry](https://www.bartleby.com/isbn_cover_images/9780395977279/9780395977279_largeCoverImage.gif)
a.
To find:Verify that the point
a.
![Check Mark](/static/check-mark.png)
Explanation of Solution
Given information:Point
Calculation:
Substitute values of x and y from point
Since, point
So, point
Substitute values of x and y from point
Since, point
So, point
Hence verified.
b.
To find:Show that the segment joining the center of circle to
b.
![Check Mark](/static/check-mark.png)
Explanation of Solution
Given information:Point
Formula used: Slope of the line
Theslope of a line passing through point
Concept used:Two lines are perpendicular if product of their slopes is
i.e. if slopes of lines are
Center of circle
Calculation:
Since, equation of circle is
So, center of circle is
Let slope of segment
Let slope of line
Here,
Now,
Since,
Hence, segment joining the center of circle to
c.
To find: Show that the segment joining the center of circle to
c.
![Check Mark](/static/check-mark.png)
Answer to Problem 38WE
The line
Explanation of Solution
Given information:Point
Concept used:A point lies on both line and circle such that the segment joining the center of circle to this point is perpendicular to the line then the line is tangent to the circle at that point.
Calculation:
From the parts (a) and (b):
Since, point
So, line
Hence, the line
Chapter 13 Solutions
McDougal Littell Jurgensen Geometry: Student Edition Geometry
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