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Circle

JEE Mathematics · 133 questions · Page 11 of 14 · Click an option or "Show Solution" to reveal answer

Q101
If the tangents at the points P\mathrm{P} and Q\mathrm{Q} on the circle x2+y2−2x+y=5x^{2}+y^{2}-2 x+y=5 meet at the point R(94,2)R\left(\dfrac{9}{4}, 2\right), then the area of the triangle PQR\mathrm{PQR} is :
A 138\dfrac{13}{8}
B 58\dfrac{5}{8}
C 54\dfrac{5}{4}
D 134\dfrac{13}{4}
Correct Answer
Option B
Solution

Equation of circle x2+y2−2x+y−5=0x^2+y^2-2 x+y-5=0 On comparing with x2+y2+2gx+2fy+c=0x^2+y^2+2 g x+2 f y+c=0

2g=−2,2f=1,c=−5g=−1,f=12,c=−5\begin{gathered} 2 g=-2,2 f=1, c=-5 \\\\ g=-1, f=\frac{1}{2}, c=-5 \end{gathered}

∴\therefore Radius of the circle

r=(−1)2+(12)2+5=52r=\sqrt{(-1)^2+\left(\frac{1}{2}\right)^2+5}=\frac{5}{2}

Length of tangent, PR=S1P R=\sqrt{S_1}

l=(94)2+(2)2−2(94)+2−5=54l=\sqrt{\left(\frac{9}{4}\right)^2+(2)^2-2\left(\frac{9}{4}\right)+2-5}=\frac{5}{4}

∴ Area of △PQR=rl3r2+l2=(52)(54)3(52)2+(54)2=52×12564254+2516=52×12564×16125=58\begin{aligned} & \therefore \text{ Area of } \triangle P Q R=\dfrac{r l^3}{r^2+l^2} \\\\ &= \dfrac{\left(\dfrac{5}{2}\right)\left(\dfrac{5}{4}\right)^3}{\left(\dfrac{5}{2}\right)^2+\left(\dfrac{5}{4}\right)^2}=\dfrac{\dfrac{5}{2} \times \dfrac{125}{64}}{\dfrac{25}{4}+\dfrac{25}{16}} \\\\ &=\dfrac{\dfrac{5}{2} \times \dfrac{125}{64} \times 16}{125}=\dfrac{5}{8}\end{aligned}

Q102
Let the locus of the midpoints of the chords of the circle x2+(y−1)2=1x^2+(y-1)^2=1 drawn from the origin intersect the line x+y=1x+y=1 at P\mathrm{P} and Q\mathrm{Q}. Then, the length of PQ\mathrm{PQ} is :
A 12\dfrac{1}{2}
B 1
C 12\dfrac{1}{\sqrt{2}}
D 2\sqrt{2}
Correct Answer
Option C
Solution

Let mid-point is (x,y)(x, y)

x2+y2−2y=0xx1+yy1−(y+y1)=x12+y12−2y1\begin{aligned} & x^2+y^2-2 y=0 \\\\ & x x_1+y y_1-\left(y+y_1\right)=x_1^2+y_1^2-2 y_1 \end{aligned}

It is passing through origin

 So, 0+0−(0+y1)=x12+y12−2y1⇒−y1=x12+y12−2y1⇒x12+y12−y1=0\begin{aligned} & \text{ So, } 0+0-\left(0+y_1\right)=x_1^2+y_1^2-2 y_1 \\\\ & \Rightarrow -y_1=x_1^2+y_1^2-2 y_1 \\\\ & \Rightarrow x_1^2+y_1^2-y_1=0 \end{aligned}
x2+y2−y=0x^2+y^2-y=0

............ (1) ∵\because It intersects the line x+y=1x+y=1 So put x=1−yx=1-y in equation (1)

(1−y)2+y2−y=02y2−3y+1=0\begin{aligned} & (1-y)^2+y^2-y=0 \\\\ & 2 y^2-3 y+1=0 \end{aligned}

⇒(y−1)(2y−1)=0⇒y=1,12∴P(0,1) and Q(12,12)\begin{aligned} & \Rightarrow (y-1)(2 y-1)=0 \\\\ & \Rightarrow y=1, \dfrac{1}{2} \\\\ & \therefore P(0,1) \text{ and } Q\left(\dfrac{1}{2}, \dfrac{1}{2}\right)\end{aligned} So, PQ=(12−0)2+(12−1)2=12P Q=\sqrt{\left(\dfrac{1}{2}-0\right)^2+\left(\dfrac{1}{2}-1\right)^2}=\dfrac{1}{\sqrt{2}}

Q103
Let C:x2+y2=4C: x^2+y^2=4 and C′:x2+y2−4λx+9=0C^{\prime}: x^2+y^2-4 \lambda x+9=0 be two circles. If the set of all values of λ\lambda so that the circles C\mathrm{C} and C\mathrm{C} intersect at two distinct points, is R−[a,b]\mathrm{R}-[\mathrm{a}, \mathrm{b}], then the point (8a+12,16 b−20)(8 \mathrm{a}+12,16 \mathrm{~b}-20) lies on the curve :
A x2+2y2−5x+6y=3x^2+2 y^2-5 x+6 y=3
B 5x2−y=−115 x^2-y=-11
C x2−4y2=7x^2-4 y^2=7
D 6x2+y2=426 x^2+y^2=42
Correct Answer
Option D
Solution

C:x2+y2=4⇒C(0,0),r1=2C′:x2+y2−4λx+9=0⇒C′(2λ,0),r2=4λ2−9∣r1−r2∣∣2−4λ2−9∣⇒4λ2>4+4λ2−9−44λ2−944λ2−9+5>0⇒λ∈R\begin{aligned} & C: x^2+y^2=4 \Rightarrow C(0,0), r_1=2 \\\\ & C^{\prime}: x^2+y^2-4 \lambda x+9=0 \Rightarrow C^{\prime}(2 \lambda, 0), r_2=\sqrt{4 \lambda^2-9} \\\\ & \left|r_1-r_2\right| \left|2-\sqrt{4 \lambda^2-9}\right| \\\\ & \Rightarrow 4 \lambda^2 > 4+4 \lambda^2-9-4 \sqrt{4 \lambda^2-9} \\\\ & 4 \sqrt{4 \lambda^2-9}+5>0 \Rightarrow \lambda \in R\end{aligned} ∣2λ∣16964λ∈(−∞,−138)∪(138,∞)λ∈R−[−138,138]a=−138,b=138⇒(8a+12,16b−20)=(−1,6)⇒6(−1)2+(6)2=42\begin{aligned} & |2 \lambda|\dfrac{169}{64} \\\\ & \lambda \in\left(-\infty, \dfrac{-13}{8}\right) \cup\left(\dfrac{13}{8}, \infty\right) \\\\ & \lambda \in R-\left[\dfrac{-13}{8}, \dfrac{13}{8}\right] \\\\ & a=\dfrac{-13}{8}, b=\dfrac{13}{8} \\\\ & \Rightarrow(8 a+12,16 b-20)=(-1,6) \\\\ & \Rightarrow 6(-1)^2+(6)^2=42\end{aligned}

Q104
Let a variable line passing through the centre of the circle x2+y2−16x−4y=0x^2+y^2-16 x-4 y=0, meet the positive co-ordinate axes at the points AA and BB. Then the minimum value of OA+OBO A+O B, where OO is the origin, is equal to
A 12
B 20
C 24
D 18
Correct Answer
Option D
Solution
(y−2)=m(x−8)⇒x-intercept ⇒(−2m+8)⇒y-intercept ⇒(−8 m+2)⇒OA+OB=−2 m+8−8 m+2f′(m)=2 m2−8=0⇒m2=14⇒m=−12⇒f(−12)=18⇒ Minimum =18\begin{aligned} & (y-2)=m(x-8) \\ & \Rightarrow x \text{-intercept } \\ & \Rightarrow\left(\frac{-2}{m}+8\right) \\ & \Rightarrow y \text{-intercept } \\ & \Rightarrow(-8 \mathrm{~m}+2) \\ & \Rightarrow \mathrm{OA}+\mathrm{OB}=\frac{-2}{\mathrm{~m}}+8-8 \mathrm{~m}+2 \\ & \mathrm{f}^{\prime}(\mathrm{m})=\frac{2}{\mathrm{~m}^2}-8=0 \\ & \Rightarrow \mathrm{m}^2=\frac{1}{4} \\ & \Rightarrow \mathrm{m}=\frac{-1}{2} \\ & \Rightarrow \mathrm{f}\left(\frac{-1}{2}\right)=18 \\ & \Rightarrow \text{ Minimum }=18 \end{aligned}
Q105
Let a circle passing through (2,0)(2,0) have its centre at the point (h,k)(\mathrm{h}, \mathrm{k}). Let (xc,yc)(x_{\mathrm{c}}, y_{\mathrm{c}}) be the point of intersection of the lines 3x+5y=13 x+5 y=1 and (2+c)x+5c2y=1(2+\mathrm{c}) x+5 \mathrm{c}^2 y=1. If h=lim⁡c→1xc\mathrm{h}=\lim \limits_{\mathrm{c} \rightarrow 1} x_{\mathrm{c}} and k=lim⁡c→1yc\mathrm{k}=\lim \limits_{\mathrm{c} \rightarrow 1} y_{\mathrm{c}}, then the equation of the circle is :
A 5x2+5y2−4x−2y−12=05 x^2+5 y^2-4 x-2 y-12=0
B 25x2+25y2−20x+2y−60=025 x^2+25 y^2-20 x+2 y-60=0
C 25x2+25y2−2x+2y−60=025 x^2+25 y^2-2 x+2 y-60=0
D 5x2+5y2−4x+2y−12=05 x^2+5 y^2-4 x+2 y-12=0
Correct Answer
Option B
Solution
3x+5y=1(2+c)x+5c2y=13c2x+5c2y=c2\begin{aligned} & 3 x+5 y=1 \\ & (2+c) x+5 c^2 y=1 \\ & 3 c^2 x+5 c^2 y=c^2 \end{aligned}

Subtracting

(2+c−3c2)x=1−c2xc=1−c22+c−3c2=(1−c)(1+c)(1−c)(3c+2)=c+13c+2y=1−3x5=1−3(c+13c+2)5=3c+2−3c−35(3c+2)yc=−15(3c+2)lim⁡c→1xc=25=hlim⁡c→1yc=−125=k\begin{aligned} & \left(2+c-3 c^2\right) x=1-c^2 \\ & x_c=\frac{1-c^2}{2+c-3 c^2}=\frac{(1-c)(1+c)}{(1-c)(3 c+2)}=\frac{c+1}{3 c+2} \\ & y=\frac{1-3 x}{5}=\frac{1-3\left(\frac{c+1}{3 c+2}\right)}{5} \\ & =\frac{3 c+2-3 c-3}{5(3 c+2)} \\ & y_c=\frac{-1}{5(3 c+2)} \\ & \lim _{c \rightarrow 1} x_c=\frac{2}{5}=h \\ & \lim _{c \rightarrow 1} y_c=\frac{-1}{25}=k \end{aligned}

Equation of circle is

25x2+25y2−20x+2y−60=025 x^2+25 y^2-20 x+2 y-60=0
Q106
A square is inscribed in the circle x2+y2−10x−6y+30=0x^2+y^2-10 x-6 y+30=0. One side of this square is parallel to y=x+3y=x+3. If (xi,yi)\left(x_i, y_i\right) are the vertices of the square, then Σ(xi2+yi2)\Sigma\left(x_i^2+y_i^2\right) is equal to:
A 152
B 148
C 156
D 160
Correct Answer
Option A
Solution

One side of square is

y=x+ky=x+k

Distance of

(5,3)(5,3)

to the line

y=x+ky=x+k

is

∣3−5−k∣2=2=∣−2−k∣=2⇒k=0 or k=−4\begin{aligned} & \frac{|3-5-k|}{\sqrt{2}}=\sqrt{2} \\ & =|-2-k|=2 \\ & \Rightarrow k=0 \text{ or } k=-4 \end{aligned}

So lines are

y=xy=x

and

y=x−4y=x-4

Now, solving these lines with circle

y=x and x2+y2−10x−6y+30=0⇒2x2−16x+30=0⇒x=3,y=3x=5,y=5y=x−4 and x2+y2−10x−6y+30=0⇒x=5,y=1x=7,y=3∑i=14xi2+yi2=9+9+25+25+25+1+49+9=152\begin{aligned} y= & x \text{ and } x^2+y^2-10 x-6 y+30=0 \\ \Rightarrow & 2 x^2-16 x+30=0 \\ \Rightarrow & x=3, y=3 \\ & x=5, y=5 \\ & y=x-4 \text{ and } x^2+y^2-10 x-6 y+30=0 \\ \Rightarrow & x=5, y=1 \\ & x=7, y=3 \\ & \sum_{i=1}^4 x_i^2+y_i^2=9+9+25+25+25+1+49+9 \\ = & 152 \end{aligned}
Q107
Let C\mathrm{C} be a circle with radius 10\sqrt{10} units and centre at the origin. Let the line x+y=2x+y=2 intersects the circle C\mathrm{C} at the points P\mathrm{P} and Q\mathrm{Q}. Let MN\mathrm{MN} be a chord of C\mathrm{C} of length 2 unit and slope −1-1. Then, a distance (in units) between the chord PQ and the chord MN\mathrm{MN} is
A 3−23-\sqrt{2}
B 2−32-\sqrt{3}
C 2−1\sqrt{2}-1
D 2+1\sqrt{2}+1
Correct Answer
Option A
Solution
 Let the line by x+y=λ\text{ Let the line by } x+y=\lambda
∴∣λ2∣=3∴λ=±32\begin{aligned} & \therefore\left|\frac{\lambda}{\sqrt{2}}\right|=3 \\ & \therefore \quad \lambda= \pm 3 \sqrt{2} \end{aligned}

∴\therefore distance between lines

x+y=2x+y=2

and

x+y=32x+y=3 \sqrt{2}

is

32−22=3−2\frac{3 \sqrt{2}-2}{\sqrt{2}}=3-\sqrt{2}
Q108
If the image of the point (−4,5)(-4,5) in the line x+2y=2x+2 y=2 lies on the circle (x+4)2+(y−3)2=r2(x+4)^2+(y-3)^2=r^2, then rr is equal to:
A 2
B 3
C 4
D 1
Correct Answer
Option A
Solution
x+41=y−52=−2(4)5⇒x=−4−85=−285,y=5−165=95∴ Image is (−285,95)\begin{aligned} & \frac{x+4}{1}=\frac{y-5}{2}=\frac{-2(4)}{5} \\ & \Rightarrow \quad x=-4-\frac{8}{5}=-\frac{28}{5}, y=5-\frac{16}{5}=\frac{9}{5} \\ & \therefore \quad \text{ Image is }\left(\frac{-28}{5}, \frac{9}{5}\right) \end{aligned}

Image lies on circle

(x+4)2+(y−3)2=r2(x+4)^2+(y-3)^2=r^2
(−285+4)2+(95−3)2=r2⇒6425+3625=r2⇒r=2\begin{aligned} & \left(\frac{-28}{5}+4\right)^2+\left(\frac{9}{5}-3\right)^2=r^2 \\ & \Rightarrow \frac{64}{25}+\frac{36}{25}=r^2 \\ & \Rightarrow r=2 \end{aligned}
Q109
Let the circles C1:(x−α)2+(y−β)2=r12C_1:(x-\alpha)^2+(y-\beta)^2=r_1^2 and C2:(x−8)2+(y−152)2=r22C_2:(x-8)^2+\left(y-\dfrac{15}{2}\right)^2=r_2^2 touch each other externally at the point (6,6)(6,6). If the point (6,6)(6,6) divides the line segment joining the centres of the circles C1C_1 and C2C_2 internally in the ratio 2:12: 1, then (α+β)+4(r12+r22)(\alpha+\beta)+4\left(r_1^2+r_2^2\right) equals
A 130
B 110
C 145
D 125
Correct Answer
Option A
Solution
C1→(x−α)2+(y−β)2=r12C2→(x−8)2+(y−152)2=r22\begin{aligned} & C_1 \rightarrow(x-\alpha)^2+(y-\beta)^2=r_1^2 \\ & C_2 \rightarrow(x-8)^2+\left(y-\frac{15}{2}\right)^2=r_2^2 \end{aligned}

Point

PP

divide the line segment internally

C1C2C_1 C_2

in the ratio 2 : 1

α×1+8×21+2=6,α=2β×1+15α×21+2=6,β=3r1=(6−2)2+(6−3)2=25=5r2=(8−6)2+(15α−6)2=52α+β+4(r12+r22)=2+3+4(52+(52)2)=5+4(1254)=130\begin{aligned} & \frac{\alpha \times 1+8 \times 2}{1+2}=6, \alpha=2 \\ & \frac{\beta \times 1+\frac{15}{\alpha} \times 2}{1+2}=6, \beta=3 \\ & r_1=\sqrt{(6-2)^2+(6-3)^2}=\sqrt{25}=5 \\ & r_2=\sqrt{(8-6)^2+\left(\frac{15}{\alpha}-6\right)^2}=\frac{5}{2} \\ & \alpha+\beta+4\left(r_1^2+r_2^2\right)=2+3+4\left(5^2+\left(\frac{5}{2}\right)^2\right) \\ & =5+4\left(\frac{125}{4}\right) \\ & =130 \\ \end{aligned}
Q110
Let a circle C of radius 1 and closer to the origin be such that the lines passing through the point (3,2)(3,2) and parallel to the coordinate axes touch it. Then the shortest distance of the circle C from the point (5,5)(5,5) is :
A 42\sqrt2
B 4
C 5
D 22\sqrt2
Correct Answer
Option B
Solution

Shortest distance of circle

CC

form

(5,5)(5,5)
=9+16−1=5−1=4\begin{aligned} & =\sqrt{9+16}-1 \\ & =5-1=4 \end{aligned}
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