Topic pack · Circles แพ็กฝึกเฉพาะหัวข้อ · วงกลม
Ten questions on one topic, ordered easy → hard. The last three are full exam difficulty. Open the mark schemes only after a real attempt. สิบข้อหัวข้อเดียว เรียงง่าย → ยาก สามข้อสุดท้ายคือระดับข้อสอบจริง เปิดมาร์คสกีมหลังลองทำจริงเท่านั้น
The circle $C$ has equation $x^2+y^2-6x+4y-12=0$. Find the centre and radius of $C$.
Mark schemeshow ▾
A circle has centre $(2,-1)$ and passes through the point $(5,3)$. Find an equation of the circle.
Mark schemeshow ▾
The points $A(1,4)$ and $B(7,-2)$ are the ends of a diameter of a circle. Find an equation of the circle.
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The circle $C$ has equation $(x-3)^2+(y+2)^2=25$. (a) Show that the point $P(6,2)$ lies on $C$. (b) Find an equation of the tangent to $C$ at $P$, giving your answer in the form $ax+by=c$ where $a,b,c$ are integers.
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(b) M1 gradient of radius $=\dfrac{2-(-2)}{6-3}=\dfrac43$ M1 tangent ⟂ radius, gradient $=-\dfrac34$ M1 $y-2=-\dfrac34(x-6)$ A1 $3x+4y=26$.
The circle $C$ has equation $x^2+y^2-4x-8y+k=0$, where $k$ is a constant. (a) Write down the coordinates of the centre of $C$. (b) Given that the radius of $C$ is $3$, find the value of $k$. (c) State the range of values of $k$ for which the equation represents a circle.
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(b) M1 $(x-2)^2+(y-4)^2=20-k$ M1 $20-k=9$ A1 $k=11$.
(c) B1 need $20-k>0$, i.e. $k<20$.
The circle $C$ has equation $x^2+y^2-2x-4y-31=0$ and $P$ is the point $(7,10)$. (a) Find the centre and radius of $C$. (b) Show that $P$ lies outside $C$. (c) Find the length of a tangent drawn from $P$ to $C$.
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(b) M1 distance $=\sqrt{(7-1)^2+(10-2)^2}=\sqrt{36+64}=10$ A1 $10>6$ so $P$ is outside $C$.
(c) M1 tangent length $=\sqrt{10^2-6^2}$ (tangent ⟂ radius, Pythagoras) A1 $=\sqrt{64}=8$.
The circle $C$ has equation $(x-4)^2+(y-3)^2=25$. The point $M(6,7)$ is the midpoint of a chord $AB$ of $C$. (a) Find an equation of the chord $AB$, giving your answer in the form $ax+by=c$. (b) Find the exact length of $AB$.
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(b) M1 $CM=\sqrt{2^2+4^2}=\sqrt{20}$ M1 half-chord $=\sqrt{25-20}=\sqrt5$ A1 $AB=2\sqrt5$.
The line $\ell$ has equation $y=2x+k$ and the circle $C$ has equation $x^2+y^2=20$. (a) Show that the $x$-coordinates of any points of intersection of $\ell$ and $C$ satisfy $5x^2+4kx+k^2-20=0$. (b) Find the values of $k$ for which $\ell$ is a tangent to $C$. (c) For the positive value of $k$, find the coordinates of the point where $\ell$ touches $C$.
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(b) M1 tangency ⟺ discriminant $=0$ A1 $(4k)^2-4(5)(k^2-20)=400-4k^2=0$ A1 $k=\pm10$.
(c) M1 repeated root $x=-\dfrac{4k}{10}=-4$ when $k=10$ A1 point $(-4,2)$.
The circle $C$ has equation $x^2+y^2-10x-8y+16=0$. (a) Find the centre and radius of $C$. (b) Show that $C$ cuts the $x$-axis at the points $A(2,0)$ and $B(8,0)$. (c) Find an equation of the tangent to $C$ at $B$, giving your answer in the form $ax+by+c=0$.
Mark schemeshow ▾
(b) M1 set $y=0$: $x^2-10x+16=0$, $(x-2)(x-8)=0$ A1 $x=2,\,8$ so $A(2,0)$, $B(8,0)$ ∎.
(c) M1 gradient of radius to $B$ $=\dfrac{0-4}{8-5}=-\dfrac43$, so tangent gradient $=\dfrac34$ A1 $y=\dfrac34(x-8)$, i.e. $3x-4y-24=0$.
The points $A(1,8)$ and $B(9,2)$ are the ends of a diameter of the circle $C$. (a) Find an equation of $C$. (b) Show that the point $D(2,1)$ lies on $C$. (c) Write down the size of angle $ADB$, giving a reason. (d) Find the area of triangle $ABD$.
Mark schemeshow ▾
(b) M1 substitute $D$: $(2-5)^2+(1-5)^2$ A1 $=9+16=25$ ✓ so $D$ lies on $C$ ∎.
(c) B1 $\angle ADB=90^\circ$ — angle in a semicircle ($AB$ is a diameter).
(d) M1 $AD=\sqrt{1^2+7^2}=\sqrt{50}$, $BD=\sqrt{7^2+1^2}=\sqrt{50}$ M1 area $=\dfrac12\,AD\cdot BD$ (legs of the right angle) A1 $=\dfrac12\cdot50=25$.