By Barry Spain, W. J. Langford, E. A. Maxwell and I. N. Sneddon (Auth.)

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This point lies on the given line and so a{a+U cos 0) + biß+it sin 0) + c = 0, from which / = -2{aa+bß+c)l(a cos θ+b sin 0). Hence the mirror image is at / _ 2(aa+bß+c) tf+6tan0 ' \a p _ 2(aa+bß+c)\ acosd+b ) ' Substituting tan 0 = b/a and simplifying, we obtain that the mirror image is at (b2-a2)a-2abß-2ca 2 2 a +b ' ~ -2aba+(a2-b2)ß-2bc\ a2+b2 y Illustration II: A line is drawn through the fixed point P(a,ß) to cut the curve x2+y2 = r2 at A and B. Show that the product PA . PB is independent of the gradient of the straight line.

Obtain the equation of the circle through the three points (1, 3), (2, — 1) and (-1,1). 4. Find the equation of the diameter of the circle x2-\-y2—2x+4y = 0 which passes through the origin. 5. Find the point which is diametrically opposite to (2, 1) on the circle x2+y2-ix+5y-4 = 0. 6. Prove that the points (9, 7) and (11, 3) lie on a circle with the origin as centre. Determine the equation of the circle. 7. Obtain the equation of the circle with centre on the x-axis and which passes through the points (1, 4) and (3, 7).

Despite this, x2+y2 = 0 does not represent two straight lines. Accordingly the vanishing of Δ is a necessary but not a sufficient condition that S = 0 represent two straight lines, called a line-pair. A further calculation yields (l1m2-l2m1)2=4(h2-ab) ; K « 2 - ^ ) 2 = 4(f2-bc (/i"2-4"i) 2 = 4(g 2 -™), ; and so it is also necessary that h2 > ab ; f2 > be ; g2 > ca. It can be shown that these inequalities are not independent when Δ=0. The complete result (which we shall not attempt to prove) is that the necessary and sufficient condition that S = 0 represent a line-pair is that Δ = 0, h2 > ab or Δ = 0, h2 = ab, f2+g2 > c{a+b).