Perpendicular Lines and the equation of a tangent
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Either paper Higher3 marks
The circle \(C\) has equation \(x^2 + y^2 = 100\)
\(O\) is the origin and \(P\) is the point \((6, -8)\), which lies on \(C\).
(a)Calculate the gradient of \(OP\).
(b)Calculate the gradient of the tangent to \(C\) at \(P\).
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Either paper Higher3 marks
The circle \(C\) has equation \(x^2 + y^2 = 625\), and \(O\) is the origin.
The point \(P(-20, -15)\) lies on \(C\).
A student says that the tangent to \(C\) at \(P\) has gradient \(-\frac{4}{3}\).
Is the student right? You must show your working.
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Either paper Higher4 marks
The circle \(C\) has equation \(x^2 + y^2 = 289\)
The point \(P(-15, 8)\) lies on \(C\).
Find the equation of the tangent to \(C\) at the point \(P\).
Give your answer in the form \(y = mx + c\).
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Either paper 3 marks
The circle \(C\) has equation \(x^2 + y^2 = 625\)
The tangent to \(C\) at the point \(P(-15, -20)\) has equation \(y = -\frac{3}{4}x - \frac{125}{4}\)
(a)Write down the coordinates of the point where the tangent crosses the \(y\)-axis.
(b)Calculate the coordinates of the point where the tangent crosses the \(x\)-axis.
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Either paper Higher4 marks
The circle \(C\) has equation \(x^2 + y^2 = 400\), and \(O\) is the origin.
The point \(P(12, -16)\) lies on \(C\). The line through \(O\) and \(P\) is continued past \(O\) until it meets \(C\) again at \(Q\).
(a)Write down the coordinates of \(Q\).
(b)Find an equation of the tangent to \(C\) at \(Q\).
Give your answer in the form \(y = mx + c\).
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Either paper Higher5 marks
The diagram shows triangle \(PQR\) on a coordinate grid, not drawn accurately.
\(P\) is on the \(y\)-axis and \(PQ\) is horizontal.
Angle \(PRQ = 90^\circ\)
\(R\) has coordinates \((-12, 10)\)
The gradient of \(PR\) is \(-\frac{1}{2}\)
Work out the coordinates of \(Q\).
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Either paper Higher4 marks
The centre of a circle is the point \(C(2, -4)\).
The point \(A(4, 3)\) lies on the circle.
Find an equation of the tangent to the circle at \(A\).
Give your answer in the form \(ax + by + c = 0\) where \(a\), \(b\) and \(c\) are integers.
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Either paper Higher4 marks
A circle has equation \(x^2 + y^2 = 1.69\)
The point \(P\) lies on the circle.
The coordinates of \(P\) are \((-1.2, 0.5)\)
The line \(l\) is the tangent to the circle at the point \(P\).
Find an equation of \(l\).
Give your answer in the form \(ax + by = c\), where \(a\), \(b\) and \(c\) are integers.
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Either paper Higher3 marks
The circle \(C\) has equation \(x^2 + y^2 = 289\)
The tangent to \(C\) at the point \(P\) is parallel to the line \(y = \frac{15}{8}x + 4\)
\(P\) lies in the second quadrant.
Find the coordinates of \(P\).
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Either paper Higher6 marks
The diagram shows a circle, centre \(O\), where \(O\) is the origin.
The point \(P(-8, -4)\) is on the circle.
The tangent to the circle at \(P\) meets the \(x\)-axis at \(A\) and the \(y\)-axis at \(B\).
(a)Show that the gradient of the tangent at \(P\) is \(-2\)
(b)Work out the length \(AB\).
Give your answer in the form \(k\sqrt{n}\), where \(k\) is an integer and \(n\) is an integer with no square factor.
You must show your working.
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(a) \(-\frac{4}{3}\)
(b) \(\frac{3}{4}\) - Yes
- \(y = \frac{15}{8}x + \frac{289}{8}\)
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(a) \((0, -\frac{125}{4})\)
(b) \((-\frac{125}{3}, 0)\) -
(a) \((-12, 16)\)
(b) \(y = \frac{3}{4}x + 25\) - \((-15, 4)\)
- \(2x + 7y - 29 = 0\)
- \(-120x + 50y = 169\)
- \((-15, 8)\)
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(a) Gradient of \(OP\) is \(\frac{ -4 }{ -8 } = \frac{1}{2}\); the tangent is perpendicular to the radius, so its gradient is \(-1 \div \frac{1}{2} = -2\)
(b) \(10\sqrt{ 5 }\)
Takes a couple of seconds the first time a sheet is made.