1)\u00a0Answer\u00a0(C)<\/strong><\/p>\nSolution:<\/b><\/p>\n
Given,<\/p>\n
Center of the circle = (0,0)<\/p>\n
Radius of the circle (r) = 6 cm<\/p>\n
$\\therefore\\ $Equation of the circle is\u00a0$x^2+y^2=r^2\\ $<\/p>\n
$=$> \u00a0$x^2+y^2=6^2\\ $<\/p>\n
$=$> \u00a0$x^2+y^2=36$<\/p>\n
$=$> \u00a0$x^2+y^2-36=0$<\/p>\n
Hence, the correct answer is Option C<\/p>\n
Question 2:\u00a0<\/b>The equation of circle with centre (1, -2) and radius 4 cm is:<\/p>\n
a)\u00a0$x^2 + y^2 + 2x – 4y = 11$<\/p>\n
b)\u00a0$x^2 + y^2 + 2x – 4y = 16$<\/p>\n
c)\u00a0$x^2 + y^2 – 2x + 4y = 16$<\/p>\n
d)\u00a0$x^2 + y^2 – 2x + 4y = 11$<\/p>\n
2)\u00a0Answer\u00a0(D)<\/strong><\/p>\nSolution:<\/b><\/p>\n
Given,<\/p>\n
Centre of the circle (a, b) =\u00a0(1, -2)<\/p>\n
Radius of the circle (r) = 4 cm<\/p>\n
$\\therefore\\ $Equation of the circle is\u00a0$\\left(x-a\\right)^2+\\left(y-b\\right)^2=r^2$<\/p>\n
$=$> \u00a0$\\left(x-1\\right)^2+\\left(y-\\left(-2\\right)\\right)^2=4^2$<\/p>\n
$=$> \u00a0$\\left(x-1\\right)^2+\\left(y+2\\right)^2=4^2$<\/p>\n
$=$> \u00a0$x^2+1^2-2.x.1+y^2+2^2+2.y.2=16$<\/p>\n
$=$> \u00a0$x^2+1-2x+y^2+4+4y=16$<\/p>\n
$=$> \u00a0$x^2-2x+y^2+4y=16-1-4$<\/p>\n
$=$> \u00a0$x^2+y^2-2x+4y=11$<\/p>\n
Hence, the correct answer is Option D<\/p>\n
Question 3:\u00a0<\/b>In $\\triangle ABC, AB = AC$. A circle drawn through B touches AC at D and intersect AB at P. If D is the mid point of AC and AP 2.5 cm, then AB is equal to:<\/p>\n
a)\u00a09 cm<\/p>\n
b)\u00a010 cm<\/p>\n
c)\u00a07.5 cm<\/p>\n
d)\u00a012.5 cm<\/p>\n
3)\u00a0Answer\u00a0(B)<\/strong><\/p>\nSolution:<\/b>
\n![](\"https:\/\/cracku.in\/media\/uploads\/screenshot.14_vpSr2ct.jpg\")
<\/p>\n
Given D is midpoint of AC so,<\/p>\n
AD = $\\frac{AC}{2}$<\/p>\n
But also given AC = AB<\/p>\n
AD = $\\frac{AB}{2}$ —-(1)<\/p>\n
AD is a tangent and APB is a secant. So the tangent secant theorem can be applied,<\/p>\n
$AD^2 = AP \\times AB$<\/p>\n
$(\\frac{AB}{2})^2 = 2.5 \\times AB$<\/p>\n
$\\frac{AB^2}{4} = 2.5 \\times AB$<\/p>\n
AB = 10 cm<\/p>\n
Question 4:\u00a0<\/b>The graph of the equations $5x – 2y + 1 = 0$ and $4y – 3x + 5 = 0$, interest at the point $P(\\alpha, \\beta)$, What is the value of $(2\\alpha – 3\\beta)$?<\/p>\n
a)\u00a04<\/p>\n
b)\u00a06<\/p>\n
c)\u00a0-4<\/p>\n
d)\u00a0-3<\/p>\n
4)\u00a0Answer\u00a0(A)<\/strong><\/p>\nSolution:<\/b><\/p>\n
$5x – 2y + 1 = 0$
\n15x – 6y + 3 = 0\u00a0—(1)
\n$3x -4y – 5 = 0$
\n15x – 20y – 25 = 0\u00a0—(2)
\nFrom eq (1) and (2),
\n14y + 28 = 0
\ny = -2
\nFrom eq(1),
\n15x + 6$\\times 2$ + 3 = 0
\nx = -1
\n$\\alpha$ = -1
\n$\\beta$ = -2
\n$(2\\alpha – 3\\beta)$
\n=\u00a0$(2 \\times (-1)\u00a0+ 3\\times 2)$ = 4<\/p>\n
Question 5:\u00a0<\/b>What is the area (in square units) of the triangular region enclosed by the graphs of the equations x + y = 3, 2x + 5y = 12 and the x-axis?<\/p>\n
a)\u00a02<\/p>\n
b)\u00a03<\/p>\n
c)\u00a04<\/p>\n
d)\u00a06<\/p>\n
5)\u00a0Answer\u00a0(B)<\/strong><\/p>\nSolution:<\/b><\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Screenshot_9_Xx4skDX.png\")
<\/figure>\nx + y = 3
\n2x + 2y = 6\u00a0—(1)
\n2x +\u00a05y = 12 —(2)
\nFrom eq (1) and eq (2),
\n3y = 6
\ny = 2
\nSo height = 2
\ny = 0 —(3)
\nput the value of y in\u00a0eq(1) and (2),
\n2x = 6
\nx = 3
\nAnd 2x = 12
\nx = 6
\nArea = $\\frac{1}{2} \\times base \\times height$
\n= $\\frac{1}{2} \\times (6 – 3) \\times 2$ = 3\u00a0square units<\/p>\n
Question 6:\u00a0<\/b>The graphs of the equations $2x + 3y = 11$ and $x – 2y + 12 = 0$ intersects at $P(x_1, y_1)$ and the graph of the equations $x – 2y + 12 = 0$ intersects the x-axis at $Q (x_2, y_2)$. What is the value of $(x_1 – x_2 + y_1 + y_2)$?<\/p>\n
a)\u00a013<\/p>\n
b)\u00a0-11<\/p>\n
c)\u00a015<\/p>\n
d)\u00a0-9<\/p>\n
6)\u00a0Answer\u00a0(C)<\/strong><\/p>\nSolution:<\/b><\/p>\n
$2x + 3y = 11$ —(1)
\n$x – 2y + 12 = 0$
\n$2x – 4y = -24$ —(2)
\nFrom eq (1) and (2),
\n7y = 35
\ny = 5 = $y_1$
\nFrom eq (1),
\n$2x + 3 $\\times$ 5 = 11$
\n2x = -4
\nx = -2 = $x_1$
\nNow,
\nThe graph of the equations $x – 2y + 12 = 0$ intersects the x-axis.
\nSo,
\n$ y = y_1$ = 0
\n$x – 0 + 12 = 0$
\nx = -12 = $x_1$
\n$(x_1 – x_2 + y_1 + y_2)$
\n= -2 + 12 + 5 + 0 = 15<\/p>\n
Question 7:\u00a0<\/b>The point of intersection of the graphs of the equations 3x \u2014 5y = 19 and 3y \u20147x + 1 =0 is P$\\left(\\alpha,\\beta\\right)$ . Whatis the value of $ \\left(3\\alpha -\\beta \\right)$ ?<\/p>\n
a)\u00a0-2<\/p>\n
b)\u00a0-1<\/p>\n
c)\u00a01<\/p>\n
d)\u00a00<\/p>\n
7)\u00a0Answer\u00a0(B)<\/strong><\/p>\nSolution:<\/b><\/p>\n
The point of intersection of the graphs of the equations 3x \u2014 5y = 19 and 3y \u20147x + 1 =0 is P$\\left(\\alpha,\\beta\\right)$
\nSo,
\n3$\\alpha \u2014 5\\beta = 19$ —(1)
\n7$\\alpha \u2014 3\\beta = 1$ —(2)
\nEq(1) multiply by 3 and eq (2) multiply by 5,
\n9$\\alpha \u2014 15\\beta = 57$ —(1)
\n35$\\alpha \u2014 15\\beta = 5$ —(2)
\nFrom eq (3) and (4),
\n26$\\alpha = -52$
\n$\\alpha = -2$
\nFrom eq (1),
\n3$\\times -2 – 5\\beta = 19$
\n$\\beta = -5$
\nNow,
\n$ \\left(3\\alpha -\\beta \\right)$
\n=\u00a0$ \\left(3\\times -2 + 5s \\right)$
\n= -1<\/p>\n
Question 8:\u00a0<\/b>The graph of the equation x \u2014 7y = \u201442, intersects the y-axis at\u00a0$P\\left(\\alpha,\\beta\\right)$ and the graph of 6x + y – 15 = 0, intersects\u00a0the x-axis at $Q\\left(\\gamma,\\delta\\right)$, What is the value of\u00a0$\\alpha+\\beta+\\gamma+\\delta?$<\/p>\n
a)\u00a0$\\frac{17}{2}$<\/p>\n
b)\u00a06<\/p>\n
c)\u00a0$\\frac{9}{2}$<\/p>\n
d)\u00a05<\/p>\n
8)\u00a0Answer\u00a0(A)<\/strong><\/p>\nSolution:<\/b><\/p>\n
The graph of the equation x \u2014 7y = \u201442, intersects the y-axis at $P\\left(\\alpha,\\beta\\right)$
\nSo, x = 0
\n0 – 7y = -42
\ny = 6
\n$\\alpha$ = 0
\n$\\beta$ = 6
\ngraph of 6x + y – 15 = 0, intersects the x-axis at $Q\\left(\\gamma,\\delta\\right)$
\nSo, y = 0
\n6x – 15 = 0
\nx = 5\/2
\n$\\gamma$ = 5\/2
\n$\\delta$ = 0
\nNow,
\n$\\alpha+\\beta+\\gamma+\\delta$
\n= 0 + 6 + 5\/2 + 0 = $\\frac{17}{2}$<\/p>\n
Question 9:\u00a0<\/b>The graphs of the equations $3x + y – 5 = 0$ and $2x – y – 5 = 0$ intersect at the point $P(\\alpha, \\beta)$. What is the value of $(3\\alpha + \\beta)$?<\/p>\n
a)\u00a04<\/p>\n
b)\u00a0-4<\/p>\n
c)\u00a03<\/p>\n
d)\u00a05<\/p>\n
9)\u00a0Answer\u00a0(D)<\/strong><\/p>\nSolution:<\/b><\/p>\n
When\u00a0 graphs of the equations intersect at the point\u00a0$P(\\alpha, \\beta)$ then,
\n$3\\alpha + \\beta – 5 = 0$ —(1)
\n$2\\alpha – \\beta – 5 = 0$ —(2),
\nOn eq(1) +\u00a0(2),
\n$5\\alpha – 10 = 0$
\n$\\alpha = 2$
\nFrom the eq(2),
\n$3 \\times 2+ \\beta – 5 = 0$
\n$\\beta = -1$
\nNow,
\n$(3\\alpha + \\beta)$ = 3 $\\times$ 2 – 1 = 6 – 1 = 5
\n$\\therefore$ The correct answer is option D.<\/p>\n
Question 10:\u00a0<\/b>The graph of x + 2y = 3 and 3x – 2y = 1 meet the Y-axis at two points having distance<\/p>\n
a)\u00a0$\\frac{8}{3}$ units<\/p>\n
b)\u00a0$\\frac{4}{3}$ units<\/p>\n
c)\u00a01 units<\/p>\n
d)\u00a02 units<\/p>\n
10)\u00a0Answer\u00a0(D)<\/strong><\/p>\nSolution:<\/b><\/p>\n
on Y axis, x=0<\/p>\n
put x = 0 in x+2y = 3<\/p>\n
2y = 3<\/p>\n
$ y = \\frac{3}{2} $<\/p>\n
putting x=0 in 3x-2y = 1<\/p>\n
-2y = 1<\/p>\n
$ \\frac{-1}{2} $<\/p>\n
therefore points on Y-axis are<\/p>\n
$ (0,\\frac{3}{2}) and (0,\\frac{-1}{2}) $<\/p>\n
required distance = $ \\sqrt ((0-0)^2 + \\sqrt (\\frac{3}{2} + \\frac{1}{2})^2 $<\/p>\n
$ = \\sqrt (0+4) $ = 2 units<\/p>\n
Question 11:\u00a0<\/b>ABCDis a cyclic quadrilateral, AB and DC when produced meet at P, if PA = 8 cm, PB = 6 cm, PC = 4 cm, then the length (in cm) of PDis<\/p>\n
a)\u00a06<\/p>\n
b)\u00a012<\/p>\n
c)\u00a08<\/p>\n
d)\u00a010<\/p>\n
11)\u00a0Answer\u00a0(B)<\/strong><\/p>\nSolution:<\/b><\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/image_8qzPOxa.png\")
<\/figure>\nGiven that,PA = 8 cm, PB = 6 cm, PC = 4 cm<\/p>\n
As per tangent & secant rule,<\/p>\n
$ PA \\times PB = PD \\times PC $<\/p>\n
=>$ PD = \\frac{8\\times6}{4}=12cm $<\/p>\n
Question 12:\u00a0<\/b>In a circle, chords AD and BC meet at a point E outside the circle. If $\\angle$BAE = 76$^\\circ$ and $\\angle$ADC= 102$^\\circ$\u00a0, then $\\angle$AEC is equal to:<\/p>\n
a)\u00a0$25^\\circ$<\/p>\n
b)\u00a0$28^\\circ$<\/p>\n
c)\u00a0$26^\\circ$<\/p>\n
d)\u00a0$24^\\circ$<\/p>\n
12)\u00a0Answer\u00a0(C)<\/strong><\/p>\nSolution:<\/b><\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/247054.png\")
<\/figure>\nIn cyclic quadrilateral ABCD, sum of opposite angles = 180$^\\circ$<\/p>\n
$=$> \u00a0$\\angle$BAE +\u00a0$\\angle$BCD =\u00a0180$^\\circ$<\/p>\n
$=$> \u00a0 76$^\\circ$ +\u00a0$\\angle$BCD = 180$^\\circ$<\/p>\n
$=$> \u00a0\u00a0$\\angle$BCD =\u00a0104$^\\circ$<\/p>\n
From the figure,<\/p>\n
$\\angle$ADC +\u00a0$\\angle$EDC =\u00a0180$^\\circ$<\/p>\n
$=$>\u00a0 102$^\\circ$ +\u00a0$\\angle$EDC = 180$^\\circ$<\/p>\n
$=$> \u00a0$\\angle$EDC =\u00a078$^\\circ$<\/p>\n
$\\angle$BCD + $\\angle$ECD = 180$^\\circ$<\/p>\n
$=$> 104$^\\circ$ + $\\angle$ECD = 180$^\\circ$<\/p>\n
$=$> $\\angle$ECD = 76$^\\circ$<\/p>\n
In\u00a0$\\triangle\\ $CDE,<\/p>\n
$\\angle$DEC +\u00a0$\\angle$ECD +\u00a0$\\angle$EDC =\u00a0180$^\\circ$<\/p>\n
$=$> \u00a0$\\angle$AEC +\u00a076$^\\circ$ +\u00a078$^\\circ$ =\u00a0180$^\\circ$<\/p>\n
$=$> \u00a0$\\angle$AEC + 154$^\\circ$ =\u00a0180$^\\circ$<\/p>\n
$=$> \u00a0$\\angle$AEC =\u00a026$^\\circ$<\/p>\n
Hence, the correct answer is Option C<\/p>\n
Question 13:\u00a0<\/b>If $ \\triangle$ABC, $\\angle$ABC=90$^\\circ$ and BD$\\bot$AC , if AD = 4cm and CD = 5cm then BD is equal to<\/p>\n
a)\u00a0$3\\sqrt{5}$<\/p>\n
b)\u00a0$2\\sqrt{5}$<\/p>\n
c)\u00a0$3\\sqrt{2}$<\/p>\n
d)\u00a0$4\\sqrt{5}$<\/p>\n
13)\u00a0Answer\u00a0(B)<\/strong><\/p>\nSolution:<\/b><\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/247050.png\")
<\/figure>\nLet\u00a0$\\angle\\ $C = x<\/p>\n
In\u00a0$ \\triangle$ABC,<\/p>\n
$\\cos x$ =\u00a0$\\frac{BC}{9}$<\/p>\n
$=$>\u00a0 BC = 9 $\\cos x$<\/p>\n
In $ \\triangle$BCD,<\/p>\n
$\\cos x$ = $\\frac{5}{BC}$<\/p>\n
$=$> \u00a0$\\cos x=\\frac{5}{9\\cos x}$<\/p>\n
$=$>\u00a0 $\\cos^2x=\\frac{5}{9}$<\/p>\n
$=$> \u00a0$\\cos x=\\frac{\\sqrt{5}}{3}$<\/p>\n
$=$> \u00a0$\\sin x=\\sqrt{1-\\cos^2x}=\\sqrt{1-\\frac{5}{9}}=\\sqrt{\\frac{4}{9}}=\\frac{2}{3}$<\/p>\n
In $ \\triangle$BCD,<\/p>\n
$\\sin x=\\frac{BD}{BC}$<\/p>\n
$=$> \u00a0$\\frac{2}{3}=\\frac{BD}{9\\cos x}$<\/p>\n
$=$> \u00a0$\\frac{2}{3}=\\frac{BD}{9\\left(\\frac{\\sqrt{5}}{3}\\right)}$<\/p>\n
$=$> \u00a0$\\frac{2}{3}=\\frac{3BD}{9\\left(\\sqrt{5}\\right)}$<\/p>\n
$=$> \u00a0BD = 2$\\sqrt{5}$<\/p>\n
Hence, the correct answer is Option B<\/p>\n
Question 14:\u00a0<\/b>In $\\triangle$ABC, $\\angle$ A= 72$^\\circ$. Its sides AB and AC are produced to the points D and E respectively. If the bisectors of the $\\angle$CBD and $\\angle$BCE meet at point O, then $\\angle$BOC is equal to:<\/p>\n
a)\u00a0$16^\\circ$<\/p>\n
b)\u00a0$54^\\circ$<\/p>\n
c)\u00a0$32^\\circ$<\/p>\n
d)\u00a0$106^\\circ$<\/p>\n
14)\u00a0Answer\u00a0(B)<\/strong><\/p>\nSolution:<\/b><\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/245732.png\")
<\/figure>\nGiven,<\/p>\n
In $\\triangle$ABC, $\\angle$A = 72$^\\circ$<\/p>\n
OB is the angular bisector of $\\angle$CBD<\/p>\n
$=$> $\\angle$OBD = $\\angle$OBC<\/p>\n
Let $\\angle$OBD = $\\angle$OBC = $x$<\/p>\n
OC is the angular bisector of $\\angle$BCE<\/p>\n
$=$> $\\angle$OCE = $\\angle$OCB<\/p>\n
Let $\\angle$OCE = $\\angle$OCB = $y$<\/p>\n
From the figure,<\/p>\n
$\\angle$ABC + $\\angle$CBD = 180$^\\circ$<\/p>\n
$=$> $\\angle$ABC + $x$ + $x$ = 180$^\\circ$<\/p>\n
$=$> $\\angle$ABC = 180$^\\circ$- 2$x$<\/p>\n
$\\angle$ACB + $\\angle$BCE = 180$^\\circ$<\/p>\n
$=$> $\\angle$ACB + $y$ + $y$ = 180$^\\circ$<\/p>\n
$=$> $\\angle$ACB = 180$^\\circ$- 2$y$<\/p>\n
In $\\triangle$ABC,<\/p>\n
$\\angle$ABC + $\\angle$ACB + $\\angle$BAC = 180$^\\circ$<\/p>\n
$=$> 180$^\\circ$- 2$x$ + 180$^\\circ$- 2$y$ + 72$^\\circ$ = 180$^\\circ$<\/p>\n
$=$> 2$x$ + 2$y$ = 180$^\\circ$ + 72$^\\circ$<\/p>\n
$=$> 2$(x+y)$ = 252$^\\circ$<\/p>\n
$=$> $x+y$ = 126$^\\circ$ ………………..(1)<\/p>\n
In $\\triangle$OBC,<\/p>\n
$\\angle$OBC + $\\angle$OCB + $\\angle$BOC = 180$^\\circ$<\/p>\n
$=$> $x$ + $y$ + $\\angle$BOC = 180$^\\circ$<\/p>\n
$=$> 126$^\\circ$ + $\\angle$BOC = 180$^\\circ$<\/p>\n
$=$> $\\angle$BOC = 180$^\\circ$- 126$^\\circ$<\/p>\n
$=$> $\\angle$BOC = 54$^\\circ$<\/p>\n
Hence, the correct answer is Option B<\/p>\n
Question 15:\u00a0<\/b>The distance between the centres of two circles of radius 2.5 cm each is 13 cm. The length (in cm)of a transverse common tangent is:<\/p>\n
a)\u00a012<\/p>\n
b)\u00a08<\/p>\n
c)\u00a06<\/p>\n
d)\u00a010<\/p>\n
15)\u00a0Answer\u00a0(A)<\/strong><\/p>\nSolution:<\/b><\/p>\n
Radius of first circle ($r_1$) = 2.5 cm<\/p>\n
Radius of second circle ($r_2$) = 2.5 cm<\/p>\n
The distance between centres of two circles ($d$) = 13 cm<\/p>\n
$\\therefore\\ $Length of the common tangent = $\\sqrt{d^2-\\left(r_1+r_2\\right)^2}$<\/p>\n
$=\\sqrt{13^2-\\left(2.5+2.5\\right)^2}$<\/p>\n
$=\\sqrt{169-25}$<\/p>\n
$=\\sqrt{144}$<\/p>\n
$=$ 12 cm<\/p>\n
Hence, the correct answer is Option A<\/p>\n
<\/p>\n","protected":false},"excerpt":{"rendered":"
CAT Coordinate Geometry Questions [Download PDF] Coordinate Geometry is one of the key topics in the CAT Quantitative Ability (QA) section. Usually, the sums in Coordinate Geometry are not tough to solve and hence one must not miss out on the questions on Coordinate Geometry. You can check out these Coordinate Geometry questions from the […]<\/p>\n","protected":false},"author":32,"featured_media":217548,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[3],"tags":[6106,5067],"class_list":{"0":"post-217537","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-cat","8":"tag-cat-2023","9":"tag-coordinate-geometry"},"better_featured_image":{"id":217548,"alt_text":"CAT Coordinate Geometry Questions PDF","caption":"CAT Coordinate Geometry Questions PDF","description":"CAT Coordinate Geometry Questions 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