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Circle

JEE Mathematics · 133 questions · Page 12 of 14 · Click an option or "Show Solution" to reveal answer

Q111
Let the circle C1:x2+y2−2(x+y)+1=0C_1: x^2+y^2-2(x+y)+1=0 and C2\mathrm{C_2} be a circle having centre at (−1,0)(-1,0) and radius 2 . If the line of the common chord of C1\mathrm{C}_1 and C2\mathrm{C}_2 intersects the y\mathrm{y}-axis at the point P\mathrm{P}, then the square of the distance of P from the centre of C1\mathrm{C_1} is:
A 4
B 6
C 2
D 1
Correct Answer
Option C
Solution
C1:x2+y2−2(x+y)+1=0C2:(x+1)2+y2=(2)2x2+y2+2x−3=0\begin{gathered} C_1: x^2+y^2-2(x+y)+1=0 \\ C_2:(x+1)^2+y^2=(2)^2 \\ x^2+y^2+2 x-3=0 \end{gathered}

Common chord is

C1−C2=0⇒2x+y−2=0\begin{aligned} & C_1-C_2=0 \\ & \Rightarrow 2 x+y-2=0 \end{aligned}

also, this line intersects the

yy

-axis at the point

P(y,0).⇒y=2\begin{aligned} & P(y, 0) . \\ & \Rightarrow y=2 \end{aligned}
P(2,0)P(2,0)

Distance of point

PP

from

(1,1)(1,1)

is

d=(2−1)2+(0−1)2=12+12d=2⇒d2=2\begin{aligned} & d=\sqrt{(2-1)^2+(0-1)^2} \\ & =\sqrt{1^2+1^2} \\ & d=\sqrt{2} \\ & \Rightarrow d^2=2 \\ \end{aligned}
Q112
Let ABCD and AEFG be squares of side 4 and 2 units, respectively. The point E is on the line segment AB and the point F is on the diagonal AC. Then the radius r of the circle passing through the point F and touching the line segments BC and CD satisfies :
A r=1\mathrm{r}=1
B 2r2−4r+1=02 \mathrm{r}^2-4 \mathrm{r}+1=0
C 2r2−8r+7=02 \mathrm{r}^2-8 \mathrm{r}+7=0
D r2−8r+8=0\mathrm{r}^2-8 \mathrm{r}+8=0
Correct Answer
Option D
Solution
CF=42−22=22=r+r2C F=4 \sqrt{2}-2 \sqrt{2}=2 \sqrt{2}=r+r \sqrt{2}
⇒(2−r)2=r⇒2=(r2−r)⇒2=r2(2−r)2⇒2(r2−4r+4)=r2⇒r2−8r+8=0\begin{aligned} & \Rightarrow \quad(2-r) \sqrt{2}=r \\ & \Rightarrow \quad \sqrt{2}=\left(\frac{r}{2-r}\right) \Rightarrow 2=\frac{r^2}{(2-r)^2} \\ & \Rightarrow 2\left(r^2-4 r+4\right)=r^2 \\ & \Rightarrow r^2-8 r+8=0 \end{aligned}
Q113
If P(6,1)\mathrm{P}(6,1) be the orthocentre of the triangle whose vertices are A(5,−2),B(8,3)\mathrm{A}(5,-2), \mathrm{B}(8,3) and C(h,k)\mathrm{C}(\mathrm{h}, \mathrm{k}), then the point C\mathrm{C} lies on the circle :
A x2+y2−74=0x^2+y^2-74=0
B x2+y2−65=0x^2+y^2-65=0
C x2+y2−61=0x^2+y^2-61=0
D x2+y2−52=0x^2+y^2-52=0
Correct Answer
Option B
Solution

Slope of

BF=(mBF)=3−18−6=1B F=\left(m_{B F}\right)=\frac{3-1}{8-6}=1
mAC=k+2h−5=−1(AC⊥BF)⇒k+2=5−h⇒k=3−h… (i) mAD=1+26−5=3mBC=k−3h−8=−13(AD⊥BC)⇒3k−9=8−h… (ii) \begin{aligned} & m_{A C}=\frac{k+2}{h-5}=-1(A C \perp B F) \\ & \Rightarrow k+2=5-h \\ & \Rightarrow k=3-h \quad \ldots \text{ (i) } \\ & m_{A D}=\frac{1+2}{6-5}=3 \\ & m_{B C}=\frac{k-3}{h-8}=-\frac{1}{3}(A D \perp B C) \\ & \Rightarrow 3 k-9=8-h \quad \ldots \text{ (ii) } \end{aligned}

From (i) & (ii)

h=−4,k=7h=-4, k=7

Now,

h2+k2=16+49=65h^2+k^2=16+49=65
h2+k2−65=0h^2+k^2-65=0

Locus is

x2+y2−65=0x^2+y^2-65=0
Q114
A circle is inscribed in an equilateral triangle of side of length 12. If the area and perimeter of any square inscribed in this circle are mm and nn, respectively, then m+n2m+n^2 is equal to
A 408
B 414
C 312
D 396
Correct Answer
Option A
Solution

Inradius of

△ABC=r=Δs=34×(12)218\triangle A B C=r=\frac{\Delta}{s}=\frac{\frac{\sqrt{3}}{4} \times(12)^2}{18}
r=23r=2 \sqrt{3}

Side length of square is

aa

, then

a2=2r2a^2=2 r^2
⇒a2=24\Rightarrow a^2=24

Area of square,

m=24m=24

Perimeter of square,

n=424n=4 \sqrt{24}
⇒m+n2=24+384=408\begin{aligned} & \Rightarrow m+n^2=24+384 \\ & =408 \end{aligned}
Q115
Let a circle C pass through the points (4, 2) and (0, 2), and its centre lie on 3x + 2y + 2 = 0. Then the length of the chord, of the circle C, whose mid-point is (1, 2), is:
A 42\sqrt{2}
B 22\sqrt{2}
C 23\sqrt{3}
D 3\sqrt{3}
Correct Answer
Option C
Solution
MAB=0⇒OM is vertical ⇒α=2∴ Centre (0)≡(2,−4)r=OA=(2−4)2+(2+4)2=40 mid point of chord is N≡(1,2)∴ON=37∴ length of chord =2r2−(ON)2=240−37=23\begin{aligned} &\begin{aligned} & \mathrm{M}_{A B}=0 \Rightarrow \mathrm{OM} \text{ is vertical } \\ & \Rightarrow \alpha=2 \\ & \therefore \text{ Centre }(0) \equiv(2,-4) \\ & \quad r=O A=\sqrt{(2-4)^2+(2+4)^2}=\sqrt{40} \end{aligned}\\ &\text{ mid point of chord is } \mathrm{N} \equiv(1,2) \quad \therefore \mathrm{ON}=\sqrt{37}\\ &\begin{aligned} \therefore \text{ length of chord } & =2 \sqrt{\mathrm{r}^2-(\mathrm{ON})^2} \\ & =2 \sqrt{40-37}=2 \sqrt{3} \end{aligned} \end{aligned}
Q116
Let the line x+y=1 meet the circle x2+y2=4x^2+y^2=4 at the points A and B. If the line perpendicular to AB and passing through the mid-point of the chord AB intersects the circle at C and D, then the area of the quadrilateral ABCD is equal to :
A 14 \sqrt{14}
B 37 3\sqrt{7}
C 214 2\sqrt{14}
D 57 5\sqrt{7}
Correct Answer
Option C
Solution

By solving x=y\mathrm{x}=\mathrm{y} with circle We get

C(2,2)D(−2,−2)\begin{aligned} & \mathrm{C}(\sqrt{2}, \sqrt{2}) \\ & \mathrm{D}(-\sqrt{2},-\sqrt{2}) \end{aligned}

By solving x+y=1\mathrm{x}+\mathrm{y}=1 with circle x2+y2=4x^2+y^2=4 we set

A(1+72,1−72)\mathrm{A}\left(\frac{1+\sqrt{7}}{2}, \frac{1-\sqrt{7}}{2}\right)

& B(1−72,1+72)\mathrm{B}\left(\dfrac{1-\sqrt{7}}{2}, \dfrac{1+\sqrt{7}}{2}\right) ∴\therefore Area of Quadrilateral ACBD =2×=2 \times Area of △BCD\triangle \mathrm{BCD}

=2×12∣2211−721+721−2−21∣=214\begin{aligned} & =2 \times \frac{1}{2}\left|\begin{array}{ccc} \sqrt{2} & \sqrt{2} & 1 \\ \frac{1-\sqrt{7}}{2} & \frac{1+\sqrt{7}}{2} & 1 \\ -\sqrt{2} & -\sqrt{2} & 1 \end{array}\right| \\ & =2 \sqrt{14} \end{aligned}
Q117
A circle C of radius 2 lies in the second quadrant and touches both the coordinate axes. Let r be the radius of a circle that has centre at the point (2,5)(2,5) and intersects the circle CC at exactly two points. If the set of all possible values of r is the interval (α,β)(\alpha, \beta), then 3β−2α3 \beta-2 \alpha is equal to :
A 10
B 12
C 14
D 15
Correct Answer
Option D
Solution
S1:(x+2)2+(y−2)2=22S_1:(x+2)^2+(y-2)^2=2^2
S2:(x−2)2+(y−5)2=r2\mathrm{S}_2:(\mathrm{x}-2)^2+(\mathrm{y}-5)^2=\mathrm{r}^2

Both circle intersect at two points

∴∣r1−r2∣<ccc2<r1+r2∣r−2∣<5<2+r⇒3<r<7r∈(3,7)α=3,β=73β−2α=15\begin{aligned} & \therefore\left|\mathrm{r}_1-\mathrm{r}_2\right|<\mathrm{c}_{\mathrm{c}} \mathrm{c}_2<\mathrm{r}_1+\mathrm{r}_2 \\ & |\mathrm{r}-2|<5<2+\mathrm{r} \\ & \Rightarrow 3<\mathrm{r}<7 \\ & \mathrm{r} \in(3,7) \\ & \alpha=3, \beta=7 \\ & 3 \beta-2 \alpha=15 \end{aligned}
Q118
Let the equation of the circle, which touches xx-axis at the point (a,0),a>0(a, 0), a>0 and cuts off an intercept of length bb on y−axisy-a x i s be x2+y2−αx+βy+γ=0x^2+y^2-\alpha x+\beta y+\gamma=0. If the circle lies below x−axisx-a x i s, then the ordered pair (2a,b2)\left(2 a, b^2\right) is equal to
A (α,β2+4γ)\left(\alpha, \beta^2+4 \gamma\right)
B (α,β2−4γ)\left(\alpha, \beta^2-4 \gamma\right)
C (γ,β2−4α)\left(\gamma, \beta^2-4 \alpha\right)
D (γ,β2+4α)\left(\gamma, \beta^2+4 \alpha\right)
Correct Answer
Option B
Solution

By pytogorus r2=a2+b24=P2\mathrm{r}^2=\mathrm{a}^2+\dfrac{\mathrm{b}^2}{4}=\mathrm{P}^2

r=4a2+b24r=\sqrt{\frac{4 a^2+b^2}{4}}

Equation of circle is (x−α)2+(y−β)2=r2(x-\alpha)^2+(y-\beta)^2=r^2

x2+y2−2ax−2py+α2+p2−r2=0 comparision x2+y2−αx+βy+r=0−α=−2a,β=−2p,r=a2⇒2a=α,4a2+b2=4p2α2+b2=4p2α2+b2=β2\begin{aligned} & x^2+y^2-2 a x-2 p y+\alpha^2+p^2-r^2=0 \\ & \text{ comparision } x^2+y^2-\alpha x+\beta y+r=0 \\ & -\alpha=-2 a, \beta=-2 p, r=a^2 \\ & \Rightarrow 2 a=\alpha, 4 a^2+b^2=4 p^2 \\ & \alpha^2+b^2=4 p^2 \\ & \alpha^2+b^2=\beta^2 \end{aligned}

So, (2a,b2)=(α,β2−4r)\left(2 a, b^2\right)=\left(\alpha, \beta^2-4 r\right)

Q119
Let C1C_1 be the circle in the third quadrant of radius 3 , that touches both coordinate axes. Let C2C_2 be the circle with centre (1,3)(1,3) that touches C1\mathrm{C}_1 externally at the point (α,β)(\alpha, \beta). If (β−α)2=mn(\beta-\alpha)^2=\dfrac{m}{n} , gcd⁡(m,n)=1\operatorname{gcd}(m, n)=1, then m+nm+n is equal to
A 22
B 13
C 9
D 31
Correct Answer
Option A
Solution
16+36=r+3⇒r=52−3⇒3−3r−r+3=α,β=9−3r3+r⇒(9−3r−3+3r)2(r+3)2⇒3612=913=mn⇒m+n=22\begin{aligned} & \sqrt{16+36}=r+3 \\ & \Rightarrow r=\sqrt{52}-3 \\ & \Rightarrow \frac{3-3 r}{-r+3}=\alpha, \beta=\frac{9-3 r}{3+r} \\ & \Rightarrow \frac{(9-3 r-3+3 r)^2}{(r+3)^2} \\ & \Rightarrow \frac{36}{12}=\frac{9}{13}=\frac{m}{n} \Rightarrow m+n=22 \end{aligned}
Q120
If the four distinct points (4,6),(−1,5),(0,0)(4,6),(-1,5),(0,0) and (k,3k)(k, 3 k) lie on a circle of radius rr, then 10k+r210 k+r^2 is equal to
A 34
B 32
C 35
D 33
Correct Answer
Option C
Solution
2r=524r2=52\begin{aligned} 2 r & =\sqrt{52} \\ 4 r^2 & =52 \end{aligned}
2r2=26r2=13 Eq n of circle is x(x−4)+y(y−6)=0k(k−4)+3k(3k−6)=0k2−4k+9k2−18k=010k2=22k10k=22∴10k+r2=35\begin{aligned} &\begin{aligned} & 2 r^2=26 \\ & r^2=13 \end{aligned}\\ &\text{ Eq }{ }^{\mathrm{n}} \text{ of circle is }\\ &\begin{aligned} & x(x-4)+y(y-6)=0 \\ & k(k-4)+3 k(3 k-6)=0 \\ & k^2-4 k+9 k^2-18 k=0 \\ & 10 k^2=22 k \\ & 10 k=22 \\ & \therefore \quad 10 k+r^2=35 \end{aligned} \end{aligned}
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