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Math  /  Geometry

QuestionThe line x+5y+22=0x+5 y+22=0 intersects the circle x2+y2+4x+8y−6=0x^{2}+y^{2}+4 x+8 y-6=0 at the point AA and BB. Find the coordinates of AA and BB. Answer a. A(7,−3),B(3,5)A(7,-3), B(3,5) b. A(7,−3),B(3,−5)A(7,-3), B(3,-5) c. A(6,−3),B(3,−7)A(6,-3), B(3,-7) d. A(6,−3),B(3,−7)A(6,-3), B(3,-7)

Studdy Solution

STEP 1

1. The line equation is x+5y+22=0x + 5y + 22 = 0.
2. The circle equation is x2+y2+4x+8y−6=0x^2 + y^2 + 4x + 8y - 6 = 0.
3. The coordinates of the intersection points AA and BB can be found by solving the system of equations formed by the line and the circle.

STEP 2

1. Express xx in terms of yy using the line equation.
2. Substitute this expression into the circle equation to form a quadratic equation in yy.
3. Solve the quadratic equation to find the yy-coordinates of points AA and BB.
4. Substitute the yy-coordinates back into the line equation to find the corresponding xx-coordinates.
5. Verify the coordinates of points AA and BB by substituting them back into the original circle equation.

STEP 3

Express xx in terms of yy using the line equation x+5y+22=0x + 5y + 22 = 0.
x=−5y−22 x = -5y - 22

STEP 4

Substitute x=−5y−22x = -5y - 22 into the circle equation x2+y2+4x+8y−6=0x^2 + y^2 + 4x + 8y - 6 = 0.
(−5y−22)2+y2+4(−5y−22)+8y−6=0 (-5y - 22)^2 + y^2 + 4(-5y - 22) + 8y - 6 = 0

STEP 5

Expand and simplify the substituted equation.
(25y2+220y+484)+y2−20y−88+8y−6=0 (25y^2 + 220y + 484) + y^2 - 20y - 88 + 8y - 6 = 0
26y2+208y+390=0 26y^2 + 208y + 390 = 0

STEP 6

Simplify the quadratic equation:
26y2+208y+390=0 26y^2 + 208y + 390 = 0
Divide the equation by 2 to simplify:
13y2+104y+195=0 13y^2 + 104y + 195 = 0

STEP 7

Solve the quadratic equation 13y2+104y+195=013y^2 + 104y + 195 = 0 using the quadratic formula y=−b±b2−4ac2ay = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} where a=13a=13, b=104b=104, and c=195c=195.
y=−104±1042−4⋅13⋅1952⋅13 y = \frac{-104 \pm \sqrt{104^2 - 4 \cdot 13 \cdot 195}}{2 \cdot 13}
y=−104±10816−1014026 y = \frac{-104 \pm \sqrt{10816 - 10140}}{26}
y=−104±67626 y = \frac{-104 \pm \sqrt{676}}{26}
y=−104±2626 y = \frac{-104 \pm 26}{26}

STEP 8

Determine the two yy-values.
y1=−104+2626=−7826=−3 y_1 = \frac{-104 + 26}{26} = \frac{-78}{26} = -3
y2=−104−2626=−13026=−5 y_2 = \frac{-104 - 26}{26} = \frac{-130}{26} = -5

STEP 9

Substitute y=−3y = -3 back into x=−5y−22x = -5y - 22.
x1=−5(−3)−22=15−22=−7 x_1 = -5(-3) - 22 = 15 - 22 = -7
Substitute y=−5y = -5 back into x=−5y−22x = -5y - 22.
x2=−5(−5)−22=25−22=3 x_2 = -5(-5) - 22 = 25 - 22 = 3

STEP 10

Verify the points (−7,−3)(-7, -3) and (3,−5)(3, -5) by substituting them back into the circle equation.
For (−7,−3)(-7, -3):
(−7)2+(−3)2+4(−7)+8(−3)−6=49+9−28−24−6=0 (-7)^2 + (-3)^2 + 4(-7) + 8(-3) - 6 = 49 + 9 - 28 - 24 - 6 = 0
For (3,−5)(3, -5):
(3)2+(−5)2+4(3)+8(−5)−6=9+25+12−40−6=0 (3)^2 + (-5)^2 + 4(3) + 8(-5) - 6 = 9 + 25 + 12 - 40 - 6 = 0
Both points satisfy the circle equation.
The coordinates of points AA and BB are:
A(−7,−3) and B(3,−5) A(-7, -3) \text{ and } B(3, -5)
Therefore, the correct answer is not among the given options indicating a miscalculation or a typo in the options. The correct points should be:
A(−7,−3) and B(3,−5) A(-7, -3) \text{ and } B(3, -5)

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