prove that the tangent at (3,-2) of the circle x²+y²=13 touches the circle x²+y²+2x-10y-26=0 and find its point of contact
Answers
QUESTION
prove that the tangent at (3,-2) of the circle x²+y²=13 touches the circle x²+y²+2x-10y-26=0 and find its point of contact
YOUR ANSWER
The circle x² + y² = 13 has its centre at (0,0) and radius sqrt(13). The slope of the line through the centre (0,0) and the tangent point (3,-2) = (-2 - 0)/(3 - 0) = -2/3 Since the tangent line to the circle and the line passing through the centre are perpendicular to each other, then the slope of the tangent line = 3/2. So the equation of the tangent line is:
The circle x² + y² = 13 has its centre at (0,0) and radius sqrt(13). The slope of the line through the centre (0,0) and the tangent point (3,-2) = (-2 - 0)/(3 - 0) = -2/3 Since the tangent line to the circle and the line passing through the centre are perpendicular to each other, then the slope of the tangent line = 3/2. So the equation of the tangent line is:(y + 2)/(x - 2) = 3/2 Simplify, you get: y = (3x - 13)/2 … (1) Now, to prove that this line is tangent to the circle of the equation x2 + y2 + 2x - 10y - 26 = 0 … (2) there must be one root of the quadratic equation obtained from solving (1) and (2) simultaneously, let us solve them by substituting (1) in (2):
The circle x² + y² = 13 has its centre at (0,0) and radius sqrt(13). The slope of the line through the centre (0,0) and the tangent point (3,-2) = (-2 - 0)/(3 - 0) = -2/3 Since the tangent line to the circle and the line passing through the centre are perpendicular to each other, then the slope of the tangent line = 3/2. So the equation of the tangent line is:(y + 2)/(x - 2) = 3/2 Simplify, you get: y = (3x - 13)/2 … (1) Now, to prove that this line is tangent to the circle of the equation x2 + y2 + 2x - 10y - 26 = 0 … (2) there must be one root of the quadratic equation obtained from solving (1) and (2) simultaneously, let us solve them by substituting (1) in (2):x^2 + (9x^2 - 78x + 169)/4 + 2x - 10[(3x - 13)/2] - 26 = 0 Simplify, you get:
The circle x² + y² = 13 has its centre at (0,0) and radius sqrt(13). The slope of the line through the centre (0,0) and the tangent point (3,-2) = (-2 - 0)/(3 - 0) = -2/3 Since the tangent line to the circle and the line passing through the centre are perpendicular to each other, then the slope of the tangent line = 3/2. So the equation of the tangent line is:(y + 2)/(x - 2) = 3/2 Simplify, you get: y = (3x - 13)/2 … (1) Now, to prove that this line is tangent to the circle of the equation x2 + y2 + 2x - 10y - 26 = 0 … (2) there must be one root of the quadratic equation obtained from solving (1) and (2) simultaneously, let us solve them by substituting (1) in (2):x^2 + (9x^2 - 78x + 169)/4 + 2x - 10[(3x - 13)/2] - 26 = 0 Simplify, you get:x^2 - 10x + 25 = 0 or (x - 5)^2 = 0 as such, x = 5 Substitute in (1) to get y:
The circle x² + y² = 13 has its centre at (0,0) and radius sqrt(13). The slope of the line through the centre (0,0) and the tangent point (3,-2) = (-2 - 0)/(3 - 0) = -2/3 Since the tangent line to the circle and the line passing through the centre are perpendicular to each other, then the slope of the tangent line = 3/2. So the equation of the tangent line is:(y + 2)/(x - 2) = 3/2 Simplify, you get: y = (3x - 13)/2 … (1) Now, to prove that this line is tangent to the circle of the equation x2 + y2 + 2x - 10y - 26 = 0 … (2) there must be one root of the quadratic equation obtained from solving (1) and (2) simultaneously, let us solve them by substituting (1) in (2):x^2 + (9x^2 - 78x + 169)/4 + 2x - 10[(3x - 13)/2] - 26 = 0 Simplify, you get:x^2 - 10x + 25 = 0 or (x - 5)^2 = 0 as such, x = 5 Substitute in (1) to get y:y = (15 - 13)/2 = 2 Therefore, the tangent point between the line y = (3x - 13)/2 and the circle x2 + y2 + 2x - 10y - 26 = 0 is (5,2)
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