CircleHard
Question
Two common tngents to the circle x2 + y2 = 2a2 and parabola y2 = 8ax are
Options
A.x = ± (y + 2a)
B.y = ± (x + 2a)
C.x = ± (y + a)
D.y = ± (x + a)
Solution
Any tangent to the parabola y2 = 8ax is
y = mx +
......(i)
If (i) is a tangent to the circle, x2 + y2 = 2a2 then, √2a = ±
⇒ m2 (1 + m2) = 2 ⇒ (m2 + 2) (m2 - 1) = 0 ; ⇒ m = ± 1
So from (i), y = ± (x + 2a)
y = mx +
......(i)If (i) is a tangent to the circle, x2 + y2 = 2a2 then, √2a = ±
⇒ m2 (1 + m2) = 2 ⇒ (m2 + 2) (m2 - 1) = 0 ; ⇒ m = ± 1
So from (i), y = ± (x + 2a)
Create a free account to view solution
View Solution FreeMore Circle Questions
For the equation x2 + y2 + 2λx + 4 = 0 which of the following can be true -...The equation of the tangents from the origin to the circle x2 + y2 + 2rx + 2hy + h2 = 0, are...Equations of circles which pass through the points (1, _2) and (3, _ 4) and touch the x-axis is...The locus of the point which moves so that the lengths of the tangents from it to two given concentric circles x2 + y2 =...The circles whose equations are x2 + y2 + c2 = 2ax and x2 + y2 + c2 − 2by = 0 will touch one another externally if...