Let a circle C : (x - h)2 + (y - k)2 = r2, k > 0, touch the x-axis at (1, 0). If the line x + y = 0 intersects the circle C at P and Q such that the length of the chord PQ is 2, then the value of h + k + r is equal to \underline\hspace2cm .

Circle PYQ — JEE Main 2022 · 24 Jun · Shift 2 · Q43 with Solution

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Correct answer: 7

Use the condition that the circle touches the x-axis at $(1,0)$

A circle

(x-h)^2+(y-k)^2=r^2

has center $(h,k)$ and radius $r$ .

Since it touches the x-axis at $(1,0)$ :

the point of contact lies vertically below/above the center, so $h=1$ ,
distance of the center from the x-axis equals the radius, so $r=k$ (and given $k>0$ ).

Thus the circle becomes

(x-1)^2+(y-k)^2=k^2.

Use the chord condition with the line $x+y=0$

The line is

x+y=0.

It cuts the circle in a chord of length $PQ=2$ .

For a circle of radius $r$ , if a line is at perpendicular distance $d$ from the center, then chord length is

2\sqrt{r^2-d^2}.

Given this equals $2$ , we get

2\sqrt{r^2-d^2}=2 \implies \sqrt{r^2-d^2}=1 \implies r^2-d^2=1.

Since $r=k$ , we need the distance from center $(1,k)$ to the line $x+y=0$ .

Find the perpendicular distance from $(1,k)$ to $x+y=0$

Write the line as

x+y+0=0.

Distance from $(x_1,y_1)$ to $Ax+By+C=0$ is

d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}.

So,

d=\frac{|1+k|}{\sqrt{2}}.

Since $k>0$ , $1+k>0$ , hence

d=\frac{1+k}{\sqrt{2}}.

Apply the chord-length equation

We have

k^2-d^2=1.

Substitute $d^2=\dfrac{(1+k)^2}{2}$ :

k^2-\frac{(1+k)^2}{2}=1.

Multiply by $2$ :

2k^2-(1+2k+k^2)=2.

So,

k^2-2k-1=2 \implies k^2-2k-3=0.

Factorizing,

(k-3)(k+1)=0.

Since $k>0$ ,

k=3.

Therefore,

r=k=3, \qquad h=1.

Compute the required sum

h+k+r=1+3+3=7.

Comparison with stored answer

Stored correct answer: $7$

Our derived answer is also $7$ , so they agree.

More from Circle

View written solutionFree

Correct answer: 7

Use the condition that the circle touches the x-axis at $(1,0)$

A circle

(x-h)^2+(y-k)^2=r^2

has center $(h,k)$ and radius $r$ .

Since it touches the x-axis at $(1,0)$ :

the point of contact lies vertically below/above the center, so $h=1$ ,
distance of the center from the x-axis equals the radius, so $r=k$ (and given $k>0$ ).

Thus the circle becomes

(x-1)^2+(y-k)^2=k^2.

Use the chord condition with the line $x+y=0$

The line is

x+y=0.

It cuts the circle in a chord of length $PQ=2$ .

For a circle of radius $r$ , if a line is at perpendicular distance $d$ from the center, then chord length is

2\sqrt{r^2-d^2}.

Given this equals $2$ , we get

2\sqrt{r^2-d^2}=2 \implies \sqrt{r^2-d^2}=1 \implies r^2-d^2=1.

Since $r=k$ , we need the distance from center $(1,k)$ to the line $x+y=0$ .

Find the perpendicular distance from $(1,k)$ to $x+y=0$

Write the line as

x+y+0=0.

Distance from $(x_1,y_1)$ to $Ax+By+C=0$ is

d=\frac{|Ax_1+By_1+C|}{\sqrt{A^2+B^2}}.

So,

d=\frac{|1+k|}{\sqrt{2}}.

Since $k>0$ , $1+k>0$ , hence

d=\frac{1+k}{\sqrt{2}}.

Apply the chord-length equation

We have

k^2-d^2=1.

Substitute $d^2=\dfrac{(1+k)^2}{2}$ :

k^2-\frac{(1+k)^2}{2}=1.

Multiply by $2$ :

2k^2-(1+2k+k^2)=2.

So,

k^2-2k-1=2 \implies k^2-2k-3=0.

Factorizing,

(k-3)(k+1)=0.

Since $k>0$ ,

k=3.

Therefore,

r=k=3, \qquad h=1.

Compute the required sum

h+k+r=1+3+3=7.

Comparison with stored answer

Stored correct answer: $7$

Our derived answer is also $7$ , so they agree.

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