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Circle A is located at (6, 5) and has a radius of 2 units. What is the equation of a line that is tangent to circle A from point C (1, 3)? y = −2.5x + 12.75 y = 6 y = 0.4x + 2.6 y = 3

User Volatility
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1 Answer

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Given:

The center of circle A is (6,5).

The radius of the circle is 2 units.

The point on the tangent is C(1,3).

Required:

We need to find the equation of the tangent.

Step-by-step explanation:

The tangent is a straight line which is of the form


y-y_1=m(x-x_1)

Where m is the slope of the tangent.


\text{ Substitute }x_1=1\text{ and }y_1=3\text{ in the equation.}


y-3=m(x-1)
y=m(x-1)+3

Consider the circle equation.


(x-h)^2+(y-k)^2=r^2

Substitute h =6, k=5, and r =2 in the formula.


(x-6)^2+(y-5)^2=2^2
(x-6)^2+(y-5)^2=4

Substitute x=1 in the equation, and we get


(4-6)^2+(y-5)^2=4
y-5=0
y=5

We get the point that lies in the circle is (4,5).

Use the points (4,5) and (6,5) to find the slope of the radius.


m=(5-5)/(6-0)=0

We know that the radius and tangent are perpendicular to each other.

The slope of the perpendicular lines is negative reciprocal.

The slope of teh tangent line is again zero.


Subsitute\text{ m =0 in the equation }y=m(x-1)+3.
y=3

Final answer:

The equation of the tangent line from point C(1,3) is


y=3.

Circle A is located at (6, 5) and has a radius of 2 units. What is the equation of-example-1
User Rayncorg
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