# 4.9 Further exercises

## Example 58

Determine the equation of the circle with centre (2, 1) and radius 3.

### Answer

The equation of the circle is

(*x* − 2)^{2} + (*y* − 1)^{2} = 3^{2},

which can be rewritten in the form

*x*^{2} + *y*^{2} − 4*x* − 2*y* − 4 = 0.

## Example 59

Determine the points of intersection of the line with equation *y* = *x* + 2 and the circle in Exercise 58.

### Answer

At the points of intersection of the line and the circle, the following equations must be satisfied:

*y* = *x* + 2

and

*x*^{2} + *y*^{2} − 4*x* − 2*y* − 4 = 0.

Substituting the expression for *y* from the first equation into the second equation, we obtain

*x*^{2} + (*x* + 2)^{2} − 4*x* − 2(*x* + 2) − 4 = 0.

Multiplying this out and collecting terms, we obtain

2*x*^{2} − 2*x* − 4 = 0,

or

*x*^{2} − *x* − 2 = 0.

This can be factorised as

(*x* − 2)(*x* + 1) = 0,

so *x* = 2 or *x* = −1.

Substituting these values of *x* into the equation *y* = *x* + 2 of the line, we find that the corresponding values of *y* are *y* = 4 and *y* = 1.

Hence the points of intersection of the line and the circle are (2, 4) and (−1, 1).

## Example 60

This question concerns the parabola *y*^{2} = 4*ax* (*a* > 0) with parametric equations *x* = *at*^{2}, *y* = 2*at* and focus *F*. Let *P* and *Q* be points on the parabola with parameters *t*_{1} and *t*_{2}, respectively. Prove that:

**(a)**if*PQ*subtends a right angle at the vertex*O*of the parabola, then*t*_{1}*t*_{2}= −4;**(b)**if*t*_{1}= 2 and*PQ*is perpendicular to*FP*, then .

### Answer

**(a)**The gradient of*OP*isand the gradient of

*OQ*is*OP*and*OQ*are perpendicular (in other words,*PQ*subtends a right angle at*O*) if*m*_{1}*m*_{2}= −1, that is, ifwhich reduces to

*t*_{1}*t*_{2}= −4, as required.**(b)**Here*P*is the point (4*a*, 4*a*). Also,*F*is (*a*, 0), so the gradient of*FP*isand the gradient of

*PQ*is*FP*and*PQ*are perpendicular if*m*_{1}*m*_{2}= −1, that is, ifwhich reduces to , so , as required.

## Example 61

This question concerns the rectangular hyperbola *xy* = *c*^{2} (*c* > 0) with parametric equations *x* = *ct*, *y* = *c*/*t*. Let *P* and *Q* be points on the hyperbola with parameters *t*_{1} and *t*_{2}, respectively.

**(a)**Determine the equation of the chord*PQ*.**(b)**Determine the coordinates of the point*N*where*PQ*meets the*x*-axis.**(c)**Determine the midpoint*M*of*PQ*.**(d)**Prove that*OM*=*MN*, where*O*is the origin.

### Answer

**(a)**The gradient of the chord*PQ*isso the equation of

*PQ*isthat is,

**(b)**At the point*N*where*PQ*meets the*x*-axis,*y*= 0 and equation (S.19) also holds, so*x*=*c*(*t*_{1}+*t*_{2}). Hence*N*is the point with coordinates (*c*(*t*_{1}+*t*_{2}), 0).**(c)**The midpoint*M*of*PQ*is**(d)**By the Distance Formula, the distance*OM*isAlso,

Thus

*OM*=*MN*.*Remark:*An alternative way of proving part (d) is as follows: Since the point*M*is vertically above the midpoint of*ON*, the triangle*OMN*must be an isosceles triangle, so*OM*=*MN*.