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َQ1 FIND THE EQUATION OF THE LINE PASSING THROGH THE POINT (-1,2)AND IS PARALLEL TO THE LINE 2X-3Y=1



Q2FIND THE EQUATION OF THE CIRCLE AT (3,-2)AND PASSES THROUGH (-1<,2) ???ANS

Q3 FIND THE CENTER AND RADUIS OF THE CIRCLE WHOSE EQUATION IS X^2+Y^2+2X-4Y-4=0 AND OBTAIN THE POINTS OF INTERSECTION OF THIS CIRCLE AND THE
LINE Y=X-1

 

 

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َQ1 FIND THE EQUATION OF THE LINE PASSING THROGH THE POINT (-1,2)AND IS PARALLEL TO THE LINE 2X-3Y=1:

The equation of any straight line in general is:

where m is the slope and

is any point on the straight line.
We have the point which is (-1,2
that is:

We need the slope m
since it is parallel to the straight line
2X-3Y=1
It has the same slope:Two parallel straight line have the same slope
therefore


Hence:




3Y -2X=8

 

 

Q2FIND THE EQUATION OF THE CIRCLE AT (3,-2)AND PASSES THROUGH (-1,2) ???ANS
the equation of a circle of center (a,b
and passing throuh the point

is:

where R is the radius of the circle
which is the distance from the center to any point on the circle Therefore the radius R here is the distance between the points
(3,-2)AND
(-1,2)

Substituting ,you will get:

Now:

 

 

Q3 FIND THE CENTER AND RADUIS OF THE CIRCLE WHOSE EQUATION IS X^2+Y^2+2X-4Y-4=0 AND OBTAIN THE POINTS OF INTERSECTION OF THIS CIRCLE AND THE
LINE Y=X-1
The general equation of the circle is:


The center is given by :



the radius is given by:


that is R= 3
To find the intersection ,Substitute the equation f the straight line in the equation of the circle:
Y = X-1

Replace Y by X - 1


Solving,you get:

Subsitute these in Y=X-1
The points of intersection are:

and

 

 

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اخوك عبدالله

 

 

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