Equation of a Circle and Tangents

The number on the right is the square of the radius.

1. Compare with x^2+y^2=r^2.
2. The radius is \sqrt{49}=7.

Square the radius in the standard equation of a circle centred at the origin.

1. Use x^2+y^2=r^2.
2. Since r^2=4^2=16, the equation is x^2+y^2=16.

Substitute the known coordinate before taking the positive square root.

1. Substitute y=6: x^2+36=100.
2. Then x^2=64.
3. Since x>0, x=8.

Substitute the two coordinates into the left side of the equation.

1. The left side is 2^2+5^2=4+25=29.
2. This equals the right side, so the point lies on the circle.

On the x-axis, y=0.

1. On the x-axis y=0, so x^2+0=36 and x^2=36.
2. Both square roots count: x=6 or x=-6.

The squared distance from the centre to the given point is the squared radius.

1. The radius is the distance from the centre O(0,0) to (7,4). By Pythagoras, r^2=7^2+4^2=49+16=65.
2. A circle with centre O has equation x^2+y^2=r^2, so the equation is x^2+y^2=65.

A tangent is perpendicular to the radius at the point of contact.

1. The radius from the origin to (0,5) is vertical.
2. The tangent is horizontal through this point, so its equation is y=5.

Work out the centre and radius of the new circle, then use Pythagoras.

1. x^2+y^2=169 has centre (0,0) and radius \sqrt{169}=13.
2. The translation moves the centre 5 to the right, to (5,0). The radius stays 13.
3. A point (0,y) on the y-axis is 5 across and y up or down from the centre. It is on the circle when 5^2+y^2=13^2 (Pythagoras).
4. y^2=169-25=144, so y=12 or y=-12.

Gradients of perpendicular lines multiply to negative one.

1. Let the tangent gradient be m.
2. Then m\times\frac{2}{3}=-1.
3. Therefore m=-\frac{3}{2}.

Find the radius gradient, then use the perpendicular gradient through the given point.

1. The radius OP goes from (0,0) to (4,7), so its gradient is \frac74.
2. A tangent is perpendicular to the radius at the point of contact, so the tangent's gradient is -\frac47.
3. Substitute P into y=-\frac47x+c: 7=-\frac{16}7+c, so c=\frac{65}7.
4. Multiply y=-\frac47x+\frac{65}7 by 7: 7y=-4x+65, so the tangent is 4x+7y=65.

The tangent at A is perpendicular to the radius CA.

1. Gradient of the radius CA: \frac{6-2}{4-1}=\frac43.
2. The tangent is perpendicular to the radius, so its gradient is -\frac34.
3. Substitute A(4,6) into y=-\frac34x+c: 6=-3+c, so c=9.
4. Multiply y=-\frac34x+9 by 4: 4y=-3x+36, so the tangent is 3x+4y=36.

Substitute y=3 into the equation of the circle.

1. Substitute y=3: x^2+9=25.
2. So x^2=16, and both square roots count: x=4 or x=-4.

Find the equation of the tangent, then put x=0.

1. The radius OP has gradient \frac5{12}.
2. The tangent is perpendicular to OP, so its gradient is -\frac{12}5: y=-\frac{12}5x+c.
3. Substitute P(12,5): 5=-\frac{144}5+c, so c=5+\frac{144}5=\frac{169}5.
4. At A, x=0, so y=c=\frac{169}5 (that is, 33.8).

At the point of contact the tangent and radius must be perpendicular.

1. The radius gradient is \frac{2}{5}.
2. Rearranging the line gives gradient -\frac{5}{2}.
3. The gradients multiply to -1, so the line is perpendicular to the radius at P and is the tangent.

Find the tangent equation and its two intercepts, then use the triangle area formula.

1. The radius OP has gradient \frac24=\frac12, so the tangent, which is perpendicular to it, has gradient -2.
2. Substitute P(4,2) into y=-2x+c: 2=-8+c, so c=10. L is y=-2x+10.
3. B: put x=0, so y=10 and B=(0,10). A: put y=0, so 2x=10 and A=(5,0).
4. Triangle OAB has a right angle at O, with OA=5 and OB=10, so its area is \frac12\times5\times10=25.

Find the missing positive coordinate before finding the tangent gradient.

1. P is on the circle, so 6^2+k^2=100. Then k^2=64, and k=8 because k>0.
2. The radius OP has gradient \frac86=\frac43, so the tangent has gradient -\frac34 (the negative reciprocal).
3. Substitute P(6,8) into y=-\frac34x+c: 8=-\frac92+c, so c=\frac{25}2.
4. Multiply y=-\frac34x+\frac{25}2 by 4: 4y=-3x+50, so the tangent is 3x+4y=50.

Use the perpendicular radius gradient to find the missing coordinate, then use its distance from the origin.

Part (a)

1. The radius gradient is 2, because 2\times(-\frac{1}{2})=-1.
2. So \frac{k}{3}=2, giving k=6.

Part (b)

1. The squared radius is 3^2+6^2=9+36=45.
2. Therefore the circle is x^2+y^2=45.

The radius OP is perpendicular to the tangent and passes through the origin.

Part (a)

1. Make y the subject: y=4x-34. The tangent has gradient 4.
2. The radius OP is perpendicular to the tangent, so it has gradient -\frac14. It passes through the origin, so it is y=-\frac14x.
3. P is on both lines. Substitute y=-\frac14x into 4x-y=34: 4x+\frac14x=34, so \frac{17}4x=34 and x=8.
4. Then y=-\frac14\times8=-2. P is (8,-2).

Part (b)

1. The radius is OP, so r^2=8^2+(-2)^2=64+4=68.
2. The circle is x^2+y^2=68.

Find both tangent equations, then solve them simultaneously.

1. The radius OA has gradient \frac23, so the tangent at A has gradient -\frac32: y=-\frac32x+c. Using A(3,2): 2=-\frac92+c, so c=\frac{13}2.
2. Multiply by 2: 2y=-3x+13, so the tangent at A is 3x+2y=13.
3. In the same way, OB has gradient -\frac23, so the tangent at B has gradient \frac32, and its equation works out as 3x-2y=13.
4. Add the two equations: 6x=26, so x=\frac{13}3. Then 3\times\frac{13}3+2y=13 gives y=0.
5. T is \left(\frac{13}3,0\right). (By symmetry, T had to be on the x-axis.)

Use the tangent equation to find the missing coordinate, then use Pythagoras.

1. The radius OP has gradient \frac36=\frac12, so the tangent has gradient -2.
2. Substitute P(6,3) into y=-2x+c: 3=-12+c, so c=15. The tangent is y=-2x+15.
3. At Q, x=8: y=-16+15=-1, so Q=(8,-1).
4. From P to Q is 2 across and 4 down. By Pythagoras, PQ^2=2^2+4^2=20.
5. PQ=\sqrt{20}=\sqrt{4\times5}=2\sqrt5.