Find the Tangent Given the Point (-4,-2 square root of 3)

Problem

x^2+y^2=28,(−4,−2√(,3))

Solution

1. Identify the equation of the circle and the given point ((x_1),(y_1))=(−4,−2√(,3))

2. Verify that the point lies on the circle by substituting the coordinates into x^2+y^2=28

(−4)^2+(−2√(,3))^2=16+12=28

3. Apply the formula for the equation of a tangent line to a circle x^2+y^2=r^2 at a point ((x_1),(y_1)) which is x*(x_1)+y*(y_1)=r^2

−4*x+(−2√(,3))*y=28

4. Simplify the equation by dividing all terms by −2 to express it in standard form.

2*x+√(,3)*y=−14

5. Isolate y to write the equation in slope-intercept form y=m*x+b

√(,3)*y=−2*x−14

y=−2/√(,3)*x−14/√(,3)

6. Rationalize the denominators if necessary.

y=−(2√(,3))/3*x−(14√(,3))/3

Final Answer

2*x+√(,3)*y+14=0

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