New chat PYQ Library Doubts Rank report

Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →

Circle question

2024 · 6 Apr · Shift 2 · Q44

Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.

PYQ Library
JEE Main
Mathematics
Circle
2024 · 6 Apr · Shift 2 · Q44

Circle question

2024 · 6 Apr · Shift 2 · Q44

JEE MainMathematicsCircleMCQ+4 / −1

If

\mathrm{P}(6,1)

be the orthocentre of the triangle whose vertices are

\mathrm{A}(5,-2), \mathrm{B}(8,3)

and

\mathrm{C}(\mathrm{h}, \mathrm{k})

, then the point

\mathrm{C}

lies on the circle :

A
$x^2+y^2-74=0$
B
$x^2+y^2-65=0$
C
$x^2+y^2-61=0$
D
$x^2+y^2-52=0$

View written solutionFree

Correct answer: B

Use the orthocentre property

If $P(6,1)$ is the orthocentre of triangle $ABC$ , then:

$AP$ is perpendicular to $BC$
$BP$ is perpendicular to $AC$

Given: $A(5,-2),\quad B(8,3),\quad C(h,k),\quad P(6,1)$

Find slope of $AP$

$m_{AP} = \frac{1-(-2)}{6-5} = \frac{3}{1} = 3$

So slope of $BC$ must be: $m_{BC} = -\frac{1}{3}$

Now, $m_{BC} = \frac{k-3}{h-8}$

Hence, $\frac{k-3}{h-8} = -\frac{1}{3}$

So, $3(k-3) = -(h-8)$ $3k-9 = -h+8$ $h+3k=17 \quad \text{(1)}$

Find slope of $BP$

$m_{BP} = \frac{1-3}{6-8} = \frac{-2}{-2} = 1$

So slope of $AC$ must be: $m_{AC} = -1$

Now, $m_{AC} = \frac{k-(-2)}{h-5} = \frac{k+2}{h-5}$

Hence, $\frac{k+2}{h-5} = -1$

So, $k+2 = -(h-5)$ $k+2 = -h+5$ $h+k=3 \quad \text{(2)}$

Solve equations (1) and (2)

From (2): $h = 3-k$

Substitute into (1): $3-k+3k=17$ $2k+3=17$ $2k=14$ $k=7$

Then, $h=3-7=-4$

So, $C=(-4,7)$

Check which circle contains $C$

For point $(-4,7)$ , $x^2+y^2 = (-4)^2+7^2=16+49=65$

Thus $C$ lies on the circle: $x^2+y^2-65=0$

Match with options

This is Option B.

Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So the derived answer agrees with the stored answer.

More from Circle

Let the circles $C_{1} : (x - α)^{2} + (y - β)^{2} = r_{1}^{2}$ and $C_{2} : (x - 8)^{2} + (y - \frac{15}{2})^{2} = r_{2}^{2}$ touch each other externally at the point $(6, 6)$ . If the point $(6, 6)$ divides the line segment joining the centres of the circles…2024 · MCQ
If the image of the point $(- 4, 5)$ in the line $x + 2 y = 2$ lies on the circle $(x + 4)^{2} + (y - 3)^{2} = r^{2}$ , then $r$ is equal to:2024 · MCQ
Let a circle passing through $(2, 0)$ have its centre at the point $(h, k)$ . Let $(x_{c}, y_{c})$ be the point of intersection of the lines $3 x + 5 y = 1$ and $(2 + c) x + 5 c^{2} y = 1$ . If $h = lim_{c \to 1} x_{c}$ …2024 · MCQ
Let the centre of a circle, passing through the points $(0, 0), (1, 0)$ and touching the circle $x^{2} + y^{2} = 9$ , be $(h, k)$ . Then for all possible values of the coordinates of the centre $(h, k), 4 (h^{2} + k^{2})$ is equal to …2024 · Numerical
Four distinct points $(2 k, 3 k), (1, 0), (0, 1)$ and $(0, 0)$ lie on a circle for $k$ equal to :2024 · MCQ
Consider a circle $(x - α)^{2} + (y - β)^{2} = 50$ , where $α, β > 0$ . If the circle touches the line $y + x = 0$ at the point $P$ , whose distance from the origin is $42$ , then $(α + β)^{2}$ is equal to …2024 · Numerical
Equations of two diameters of a circle are $2 x - 3 y = 5$ and $3 x - 4 y = 7$ . The line joining the points $(- \frac{22}{7}, - 4)$ and $(- \frac{1}{7}, 3)$ intersects the circle at only one point $P (α, β)$ . Then, $17 β - α$ …2024 · Numerical
If the circles $(x + 1)^{2} + (y + 2)^{2} = r^{2}$ and $x^{2} + y^{2} - 4 x - 4 y + 4 = 0$ intersect at exactly two distinct points, then2024 · MCQ