
To calculate:The equation for the circle and line in the given figure.

Answer to Problem 129RE
The equation for the circle and line in the figure are x2+y2=169 and 5x−12y+169=0 respectively.
Explanation of Solution
Given information:
The point P(−5,12) lying on the tangent to the circle with center (h,k)=(0,0)
Formula used:
For a given circle with center (h,k) and radius r the equation is given as:
(x−h)2+(y−k)2=r2
This is referred to as the Standard form for the equation of a given circle.
The distance D between two points say (x1,y1) and (x2,y2) is given as:
D=√(x2−x1)2+(y2−y1)2
Slope m of the line passing through two points in general say P1=(x1,y1) and P2=(x2,y2) is:
m=y2−y1x2−x1
Slope-intercept equation for a given line which has slope as m and y −intercept as b is:
y=mx+b
Two-intercept equation for a given line which has x -intercept as a and y −intercept as b is:
xa+yb=1
When two lines are perpendicular then the product of their slopes is zero that is m1.m2=−1
When two lines are parallel then their slope are equal that is m1=m2
Calculation:
From the given figure it is clear that center of the circle is (h,k)=(0,0)
Recall for a given circle with center (h,k) and radius r the equation is given as:
(x−h)2+(y−k)2=r2
Therefore, replacing the values we get:
(x−0)2+(y−0)2=r2
x2+y2=r2 (1)
Now to find radius r which is the distance from the center (h,k)=(0,0) to the point P(−5,12)
Recall, the distance D between two points say (x1,y1) and (x2,y2) is given as:
D=√(x2−x1)2+(y2−y1)2
Before applying distance formula the following must be known:
x1=0y1=0x2=−5y2=12
Therefore,
D=r=√(−5−0)2+(12−0)2=√25+144=√169=13
Hence radius r=13
Put this value in (1)
Therefore, the required equation for circle is:
x2+y2=132x2+y2=169
Hence, the equation for the given circle is x2+y2=169
Now for the equation of line:
Recall, slope m of the line passing through two points in general say P1=(x1,y1) and P2=(x2,y2) is:
m=y2−y1x2−x1
As it is known:
x1=0y1=0x2=−5y2=12
Therefore, slope of radius for above values is:
m1=12−0−5−0m1=−125
Now as the tangent is perpendicular to the circle, therefore the product of their slopes is −1
Slope of line is m2=512
And it passes through P(−5,12) that is x=−5y=12
Put these values in the general equation for the line which is y=m2x+b (2)
Which gives:
12=512(−5)+bb=12+2512b=144+2512b=16912
Now put m2=512 and b=16912 in (2) to get:
y=512x+1691212y=5x+1695x−12y+169=0
Thus, required equation of line is 5x−12y+169=0
Chapter 1 Solutions
EBK PRECALCULUS: MATHEMATICS FOR CALCUL
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