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sashaice [31]
3 years ago
9

If x^2 + y^2 = 25, find the equation of the line tangent to the circle when x = -4

Mathematics
1 answer:
bekas [8.4K]3 years ago
7 0

Answer:

In the point (-4,3) tangent is : y=4/3x+25/3 i.e, y=1.33x+8.33

In the point (-4,-3) tangent is: y=-4/3x-25/3 i.e, y=-1.33x-8.33

Step-by-step explanation:

The center of the circle is in (0,0), because x^2+y^2=(x-0)^2+(y-0)^2=25. And the radius is 5, because 5^2=25.

We have first center: (x1,y1)=(0,0)

The point of the tangent is in x=-4, so x^2+y^2=25 is 16+y^2=25,i.e y=3 and y=-3. Like you can see we have 2 tangents to the circle when x=-4. the point (-4,3) and (-4,-3).

We will do for the first point:

The point of tangent is (x2,y2)=(-4,3)

The gradient  of the radius

mr=(y2-y1)/(x2-x1)=(3-0)/(-4-0)=-3/4

The gradient of the tangent mt is:

mr*mt=-1 (they are ortogon)

mt=-1/mr=-1/(-3/4)=4/3

the equation of the line trought one point is:

y-y2=k(x-x2), where k is mt,i.e. k=4/3

y-3=4/3(x-(-4))

y=4/3x+16/3+3

y=4/3x+25/3

y=1.33x+8.33

Now for the 2nd point (-4,-3),

mr=-3/-4=3/4

mt=-4/3

y+3=-4/3(x-(-4))

y=-4/3x-16/3-3

y=-4/3x-25/3

y=-1.33x-8.33

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