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Q.No- 39) From a point P(1,0) , a tangent PA is drawn to the circle x 2 +y 2 -8x+12=0 , A being the point of contact. Find the equations of the tangents to the circle from the middle point of PA if A is in the first quadrant. 41) A tangent is drawn to the circle (x-a) 2 + y 2 = b 2 and a perpendicular tangent to the circle (x+a) 2 + y 2 = c 2 . Find the locus of their intersection and prove that the bisectors of the angles between them always touch one or other of two other fixed circles. 90) The base of a triangle through a fixed point (a,b) and its sides are respectively bisected at right angles by the lines y 2 -8xy-9x 2 =0. Prove that the locus of the vertex is a circle. Find its Equation.
39. Dear Student, Please have a look at the symmetry of the problem. First let us find the coordinates of A(x, y). One equation is – (4–1)2 + (0–1)2 = (x–1)2 + (y–0)2 + (4–x)2 + (0–y)2 …. (i) Second equation is – Slope of PA × slope of AL + –1 ….. (ii) Solving these two equations you get the coordinates of A(x, y) Then by applying mid-point formula, we get the coordinates of mid-point of PA say (x1, y1) Then we get the equation of pair of tangents to the circle from (x1, y1) as ?S1 = T2 Where S = x2 + y2 + 2fx + 2fy + c S1 = S12 + y12 + 2fx1 + 2fy1 + c and T = x x1 + y y1 + y(x + x1) + f(y + y1) + c
39. Dear Student,
Please have a look at the symmetry of the problem.
First let us find the coordinates of A(x, y).
One equation is –
(4–1)2 + (0–1)2 = (x–1)2 + (y–0)2 + (4–x)2 + (0–y)2 …. (i)
Second equation is –
Slope of PA × slope of AL + –1 ….. (ii)
Solving these two equations you get the coordinates of A(x, y)
Then by applying mid-point formula, we get the coordinates of mid-point of PA say (x1, y1)
Then we get the equation of pair of tangents to the circle from (x1, y1) as
?S1 = T2
Where S = x2 + y2 + 2fx + 2fy + c
S1 = S12 + y12 + 2fx1 + 2fy1 + c
and T = x x1 + y y1 + y(x + x1) + f(y + y1) + c
ANSWER 39
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