Two Dimensional Geometry jee mains questions | JEE Main Two Dimensional Geometry Previous Year Question

Q 41 :

The set of all values of $a^{2}$ for which the line $x + y = 0$ bisects two distinct chords drawn from a point $P (\frac{1 + a}{2}, \frac{1 - a}{2})$ on the circle $2 x^{2} + 2 y^{2} - (1 + a) x - (1 - a) y = 0,$ is equal to [2023]

$(8, \infty)$
$(0, 4]$
$(4, \infty)$
$(2, 12]$

1

2

3

4

(1)

Q 42 :

Let the point $(p, p + 1)$ lie inside the region $E = {(x, y) : 3 - x \leq y \leq \sqrt{9 - x^{2}}, 0 \leq x \leq 3} .$ If the set of all values of $p$ is the interval $(a, b),$ then $b^{2} + b - a^{2}$ is equal to _____ . [2023]

0%

(3)

$Given, 3 - x \leq y \leq \sqrt{9 - x^{2}}$

$3 - x \leq y \Rightarrow x + y - 3 \geq 0 \dots (i)$

$y \leq \sqrt{9 - x^{2}} \Rightarrow x^{2} + y^{2} \leq 9$

$Now, (p, p + 1) lies inside the region E .$ $∴$ $p + p + 1 - 3 \geq 0$

$\Rightarrow p \geq 1 and p^{2} + {(p + 1)}^{2} \leq 9$

$\Rightarrow 2 p^{2} + 2 p - 8 \leq 0 \Rightarrow p^{2} + p - 4 \leq 0$

$\Rightarrow p \in (- \frac{(1 + \sqrt{17})}{2}, \frac{\sqrt{17} - 1}{2})$

$∴$ $p \in (1, \frac{\sqrt{17} - 1}{2}) (as p \geq 1)$

$a = 1, b = \frac{\sqrt{17} - 1}{2} \Rightarrow b^{2} + b - a^{2} = 3$

Q 43 :

A circle passing through the point $P (α, β)$ in the first quadrant touches the two coordinate axes at the points A and B. The point P is above the line AB. The point Q on the line segment AB is the foot of perpendicular from P on AB. If PQ is equal to 11 units, then the value of $α β$ is ________. [2023]

0%

(121)

$Let the equation of the circle be$

${(x - a)}^{2} + {(y - a)}^{2} = a^{2},$

It passes through $P (α, β)$

$∴$ ${(α - a)}^{2} + {(β - a)}^{2} = a^{2}$

$\Rightarrow α^{2} + β^{2} - 2 a α - 2 a β + a^{2} = 0$ ...(i)

Here, the equation of line AB is $x + y = a$

Let $Q (α', β')$ be the foot of the perpendicular from $P$ on AB

$∴$ $\frac{α' - α}{1} = \frac{β' - β}{1} = \frac{- (α + β - a)}{2}$

${(P Q)}^{2} = {(α' - α)}^{2} + {(β' - β)}^{2} = \frac{1}{4} {(α + β - a)}^{2} + \frac{1}{4} {(α + β - a)}^{2}$

$\Rightarrow {(11)}^{2} = \frac{1}{2} {(α + β - a)}^{2}$

$\Rightarrow 242 = α^{2} + β^{2} + a^{2} + 2 α β - 2 a α - 2 a β$

$\Rightarrow 242 = 2 α β$ [From (i)]

$\Rightarrow α β = 121$

Q 44 :

Consider a circle $C_{1} : x^{2} + y^{2} - 4 x - 2 y = α - 5$ . Let its mirror image in the line $y = 2 x + 1$ be another circle $C_{2} : 5 x^{2} + 5 y^{2} - 10 f x - 10 g y + 36 = 0$ . Let $r$ be the radius of $C_{2}$ . Then $α + r$ is equal to _______ . [2023]

0%

(2)

We have,

$C_{1} : x^{2} + y^{2} - 4 x - 2 y - (α - 5) = 0$

$∴$ $Centre is (2, 1), r = \sqrt{4 + 1 + α - 5} = \sqrt{α}$

$C_{2} : 5 x^{2} + 5 y^{2} - 10 f x - 10 g y + 36 = 0$

$i.e., x^{2} + y^{2} - 2 f x - 2 g y + \frac{36}{5} = 0$

$Centre is (f, g), r = \sqrt{f^{2} + g^{2} - \frac{36}{5}}$

$The image of (2, 1) with respect to 2 x - y + 1 = 0 is (f, g),$

$then \frac{f - 2}{2} = \frac{g - 1}{- 1} = \frac{- 2 (4 - 1 + 1)}{5}$

$\Rightarrow \frac{f - 2}{2} = \frac{- 8}{5} and g - 1 = \frac{8}{5}$

$\Rightarrow f - 2 = - \frac{16}{5} and g - 1 = \frac{8}{5}$

$\Rightarrow f = 2 - \frac{16}{5} = \frac{- 6}{5}$ and $g = \frac{13}{5}$

$So, (f, g) = (- \frac{6}{5}, \frac{13}{5})$

$and r = \sqrt{\frac{36}{25} + \frac{169}{25} - \frac{36}{5}} \Rightarrow r = 1$

$\Rightarrow r = 1$

$∴$ $α + r = 1 + 1 = 2$

Q 45 :

Two circles in the first quadrant of radii $r_{1}$ and $r_{2}$ touch the coordinate axes. Each of them cuts off an intercept of 2 units with the line $x + y = 2$ . Then $r_{1}^{2} + r_{2}^{2} - r_{1} r_{2}$ is equal to _______ . [2023]

0%

(7)

Q 46 :

Points P(-3, 2), Q(9, 10) and R( $α$ , 4) lie on a circle C with PR as its diameter. The tangents to C at the points Q and R intersect at the point S. If S lies on the line $2 x - k y = 1$ , then $k$ is equal to ______ . [2023]

0%

(3)

$Since P Q ⊥ Q R, so m_{P Q} \cdot m_{Q R} = - 1$

$\Rightarrow \frac{10 - 2}{9 + 3} \times \frac{10 - 4}{9 - α} = - 1 \Rightarrow α = 13$

$Since Q O ⊥ Q S, so m_{O Q} \cdot m_{Q S} = - 1 \Rightarrow m_{Q S} = - \frac{4}{7}$

$Equation of Q S : y - 10 = - \frac{4}{7} (x - 9) \Rightarrow 4 x + 7 y = 106 (i)$

$Since O R ⊥ S R$

$m_{O R} \cdot m_{R S} = - 1$

$\Rightarrow m_{R S} = - 8$

$Equation of R S : y - 4 = - 8 (x - 13)$

$\Rightarrow 8 x + y = 108 (ii)$

$Solving (i) and (ii)$

$x = \frac{25}{2}, y = 8$

$Since S (x, y) lies on the line 2 x - k y = 1$

$25 - 8 k = 1 \Rightarrow 8 k = 24 \Rightarrow k = 3$

Q 47 :

A circle with centre (2, 3) and radius 4 intersects the line $x + y = 3$ at the points P and Q. If the tangents at P and Q intersect at the point $S (α, β)$ , then $4 α - 7 β$ is equal to ______ . [2023]

0%

(11)

$The equation of circle can be written as,$

${(x - 2)}^{2} + {(y - 3)}^{2} = 4^{2} \Rightarrow x^{2} + y^{2} - 4 x - 6 y - 3 = 0$

$Chord of contact at (α, β) is$

$\Rightarrow α x + β y - 2 (x + α) - 3 (y + β) - 3 = 0$

$\Rightarrow (α - 2) x + (β - 3) y - (2 α + 3 β + 3) = 0 ...(i)$

$∵$ $But the equation of chord of contact is given:$

$x + y - 3 = 0 ...(ii)$

$Comparing equations (i) and (ii):$

$\frac{α - 2}{1} = \frac{β - 3}{1} = - (\frac{2 α + 3 β + 3}{- 3})$

$On solving, α = - 6, β = - 5$

$So, 4 α - 7 β = 4 (- 6) - 7 (- 5) = 11$

Q 48 :

Let $P (a_{1}, b_{1})$ and $Q (a_{2}, b_{2})$ be two distinct points on a circle with center $C (\sqrt{2}, \sqrt{3})$ . Let O be the origin and OC be perpendicular to both CP and CQ. If the area of the triangle OCP is $\frac{\sqrt{35}}{2}$ , then $a_{1}^{2} + a_{2}^{2} + b_{1}^{2} + b_{2}^{2}$ is equal to ________ . [2023]

0%

(24)

$Area of ∆ O C P = \frac{\sqrt{35}}{2}$

$\frac{1}{2} \times P C \times O C = \frac{\sqrt{35}}{2}$

$\frac{1}{2} \times P C \times \sqrt{5} = \frac{\sqrt{35}}{2}$

$P C = \sqrt{7}$

$Now, O Q^{2} = 5 + 7 = 12$

$O P^{2} = 5 + 7 = 12$

$a_{1}^{2} + b_{1}^{2} = O P^{2} and a_{2}^{2} + b_{2}^{2} = O Q^{2}$

$∴$ $a_{1}^{2} + a_{2}^{2} + b_{1}^{2} + b_{2}^{2} = O P^{2} + O Q^{2} = 12 + 12 = 24$

Topic Question Set